StudySteps - WordPress.com · used to describe the outcomes of an experiment. In this case ‘S 1 ’ provides more information than ‘S 2 ’. If we know which element in S 1 occurs,

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1. TERMS AND DEFINITIONS USED IN PROBABILITY :

1.1 Random ExperimentAn experiment is called random if(i) all the outcomes of the experiment are known in advance(ii) the exact outcome of any specific performance of the experiment is unpredicatable i.e. not

known in advance. For Example drawing a card from a well shuffled pack of 52 playingcards is a random experiment.

1.2 Sample SpaceA set whose elements represent all possible outcomes of a random experiment is called the samplespace and is usually represented by ‘S’.Consider the experiment of tossing a die. If we are interested in the number on the top face, thensample space would be S

1 = {1, 2, 3, 4, 5, 6}. If we are interested only in whether the number is

even or odd, then sample space is S2 = {even, odd}. Clearly more than one sample space can be

used to describe the outcomes of an experiment. In this case ‘S1’ provides more information than

‘S2’. If we know which element in S

1 occurs, we can tell which outcome in S

2 occurs; however,

knowledge of what happens in S2 in no way helps us to know which element in S

1 occurs.

In general it is desirable to use a sample space that gives the maximum information concerning theoutcomes of the experiment.

1.3 Sample PointEach element of the sample space is called a sample point.

1.4 EventAn event is a subset of the sample space. When a die is rolled, sample space is S = {1, 2, 3, 4, 5, 6}.Let A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3, 4}. Here A is the event of occurrence of an odd number,B is the event of occurrence of an even number and C is the event of occurrence of a number lessthan 5.

1.5 Simple EventAn event is called a simple event, if it is a singleton subset of the sample space S.

1.6 Compound EventA subset of the sample space S which contains more than one element is called a compound event

1.7 Equally likely EventsA set of events is said to be equally likely if taking into consideration all the relevant factors there isno reason to expect one of them in preference to the others. For example when a fair coin istossed, the occurrence of a tail or a head are equally likely.

1.8 Exhaustive EventsA system of events is said to be exhaustive if on each performance of the experiment at least one ofthe events of the system is must to occur. In set theoretic notation, events A

1, A

2, . . . , A

m are

exhausitve ifm

ii 1

A S

. For example on throwing of a die, the events {1, 2}, {2, 3, 4}, {5} and

{4, 5, 6} form an exhaustive system of events.

Theory- Probability

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1.9 Mutually Exclusive EventsA set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrenceof any of the remaining events. For example, when we throw a pair of dice, the events “ a sum of 5occurs”, “a sum of 7 occurs” and “a sum of 9 occurs” are mutually exclusive. In set theoretic notation,

events A1, A

2, . . . , A

m are mutually exclusive if i jA A for i j and 1 i, j m .

2. CLASSICAL DEFINITION OF PROBABILITYIf there are n exhausitve mutually exclusive and equally likely out comes of an experiment and m of

them are favourable to an event A, then the probability of the happening of A is equal tomn

and it is

denoted by P(A). Clearly P(A) is a non-negative number not greater than unity. So 0 P(A) 1 .If probability of happening of an event A is 1, then A is called certain event and if probability ofhappening of event A is zero, then A is called impossible event.

Illustration 1 :Six boys and six girls sit in a row randomly, find the probability that all the 6 girls sit together.

Solution:Let S be the sample space and E the event that all the 6 girls sit together . Then n(S) = Total numberof ways of seating 6 boys and 6 girls in a row = 12! and n(E) = number of ways of seating 6 boys and6 girls in a row so that all the 6 girls sit together = 7!6!

P(E) =

n E 6!7! 112! 132n S .

Drill Exercise - 1

1. Two numbers are selected at random from 1, 2, 3, ......, 100 and are multiplied. Find theprobability correct to two places of decimals that the product thus obtained, is divisibleby 3.

2. Each coefficient in the equation ax2 + bx + c = 0 is determined by throwing an ordinary die. Findthe probability that the equation will have equal roots.

3. Four cards are drawn from a pack of 52 playing cards. Find the probability (correct upto twoplaces of decimals) of drawing exactly one pair.

4. From a bag containing 5 red different pairs and 4 white different pairs of socks. Find the probabilityof getting a pair by drawing 2 socks.

5. Seven white balls and three black balls are randomly placed in a row. Find the probability that no twoblack balls are placed adjacently.

2.1 Odds in Favour and Odds Against an EventAs a result of an experiment, if p of the outcomes are favourable to an event E and q of the outcomesare against it, then we say that odds are p to q in favour of E or odds are q to p


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against E.

Odds in favour of an event E =number of favourable cases p

number of unfavourable cases q

and odds against an event E =number of unfavourable cases q

number of favourable cases p

If odds in favour of an event E are p : q then the probability of the occurrence of that event isp

p q

Similarly the probability of the non-occurrence of that event isq

p q

Illustration 2 :What are the odds in favour of throwing at least 8 in a single throw with two dice?

Solution :Let S be the sample space and E be the event of occurrence of a total of at least 8, then

n(S) = 6 6 36 and n(E) = 15

P(E) =1536

15 21P E 1 P(E) 1

36 36

Odds infavour of E = P(E) : P E = 5 : 7 ]

drill exercise - 2

1. In a given race, the odds in favor of four horses A, B, C and D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6respectively. Assuming that a dead heat is impossible, find the chance that one of them wins the race.

2. The chance of one event happening is the square of the chance of a 2nd event, but odds against thefirst are the cubes of the odds against the 2nd . Find the chances of each.

3. Three critics review a book. Odds in favour of the book are 5 : 2, 4 : 3 and 3 : 4 respectively for thethree critics. Find the probability that majority are in favour of the book.

4. Suppose that it is 9 to 7 against a person A who is now 35 years of age living till he is 65 and 3to 2 against a person B now 45 living till he is 75, then find the chance that one atleast of thesepersons will be alive 30 years hence.

5. What are the odds in favours of getting a spade if the card is drawn from a well-shuffled deck ofcards ? What are the odds in favour of getting a king ?

3.1 SET THEORETIC PRINCIPLESIf ‘A’ and ‘B’ be any two events of the sample space, then A B would stand for occurrence of

atleast one of them and AB stands for simultaneous occurrence of A and B. A or A stands fornon–occurrence of A.A B (or (A B) stands for non–occurrence of both A and B. A B stands for ‘the occurrence


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of B implies the occurrence of A.If A and B are any two events, then P(AB) = P(A) + P(B) – P (AB). If A and B are mutuallyexclusive, P(AB) = P(A) + P(B).

Hence P A = P(A) = 1 – P(A)P(AB) = P(A) – P(AB)Now P(exactly one of A and B occurs)

= P A B P A B = P(A) + P(B) 2P(A B) = P(A B) P(A B)

3.2 Some Theorems

(i) If A B , then (i) P A P B and (ii) P(B – A) = P(B) – P(A)

(ii) n n

i ii 1i 1

P A P A

and equality holds if and only if events AA

i, i = 1, 2, . . . , n are

exclusive.

(iii)n

ii 1

P A 1

and equality holds if and only if events AA

i, i = 1, 2, . . . n are exhaustive.

(iv) n

ii 1

P A = 1 if events AA

i, i = 1, 2, . . ., n are exclusive and exhaustive.

Note : If events A1, A

2, . . . , A

n are exclusive, then

n n

i ii 1i 1

P A P A

. This is called the rule of

sum.

3.3 Some Useful Formulas(i) For two events A and B

(a) P(at least one out of them) = P(A) + P(B) – )BA(P (b) P (exactly one out of them) = P(A) + P(B) – 2 )BA(P

(ii) For three events A, B and C(a) P(at least one out of them) =

P A P B P C P A B P B C P C A P A B C (b) P (at least two out of them) =

)CBA(P2)BA(P)AC(P)CB(P (c) P (exactly two out of them) =

)CBA(P3)BA(P)AC(P)CB(P (d) P (exactly one out of them) =

)CBA(P3)BA(P2)AC(P2)CB(P2)C(P)B(P)A(P Illustration 3 :

From the set S = {1, 2, 3, . . . 3n}, three numbers are chosen at random. Find the probability thatthe sum of the chosen numbers is divisible by 3.

Solution :Let S

kS, having those numbers, which leave a remainder of k, k = 0, 1, 2, when divided by 3.Obviously n(S

k) = n.


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Let Ek be the event that the numbers are chosen from S

k, k = 0, 1, 2. Let E

3 be the event that the

chosen numbers are from the sets S0, S

1 and S

2 (one number from each set). Let E be event that

the sum of the chosen numbers is divisible by 3. Now P(E0) = P(E

1) = P(E

3) =

n3

3n3

C

C.

P(E3) = 3n

3

n n n

C

.

The required probability = P(E) = P 0 1 2 3E E E E

= P(E0) + P(E

1) + P(E

2) + P(E

3) (using rule of sum )

=n 3

33n

3

3. C n

C

.

Drill Exercise -3

1. Let A and B be two events defined on a sample space. Given P(A) = 0.4, P(B) = 0.80 and P A B = 0.10.

Find (i) P A B (ii) P A B A B

2. Let A and B be two events with P(A) =1

2, P(B) =

1

3and P 1

A B4

. Find

(i) PA

B

(ii) PB

A

(iii) P A B (iv) PA

B

(v) PB

A

3. If two events A and B are such that P(Ac) = 0.3, P(B) = 0.4 and P(ABc) = 0.5 then

find the value of ]BA/(B[P c .

4. The probabilities of three events A, B and C are P(A) = 0.6, P(B) = 0.4, P(C) = 0.5. If

2.0)CBA(P,3.0)CA(P,8.0)BA(P and 85.0)CBA(P . Find

85.0)CBA(P . Find )CB(P .

5. A and B are two independent events. The probability that both occur simultaneously is 1/6 and theprobability that neither occurs is 1/3. Find the probabilities of occurrence of the events A and Bseparately.

4. INDEPENDENT EXPERIMENTSIf two random experiments are performed and their outcomes are independent of each other, then theexperiments are called independent experiments.

For example(i) Consider the tossing of a coin twice, clearly the outcome for the second toss is not effected by theresult of the first toss. So the two tosses are independent random experiments.


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(ii) Consider the drawing of two balls one after the other, with replacement, from an urn containingtwo or more balls. Then the two draws are independent of each other. So considering each draw asan experiment, the two experiments are independent random experiments.

(iii) In throwing of a die and a coin together or one after the other are independent experiments.

Remarks(i) In drawing of two cards, without replacement, from a well-shuffled ordinary pack of 52-

playing cards, the two draws are not independent experiments.

(ii) Let E1 and E

2 be two independent random experiments.Let A be an event of experiment

E1 only and B be an event of experiment E

2 only. Then P(A occurs in E

1 and B occurs in

E2) = P(A). P(B).

For example if a coin and a die are thrown together and A = {H}, B = {1, 2, 3, 4}, then P(head on

coin and a number 4 on die) = P(A). P(B) =1 4 12 6 3 .

5. BINOMIAL DISTRIBUTION FOR SUCCESSIVE EVENTSIf probability of occuring an event is P and it remains P in each trial of the experiment. Now if experimentis repeated ‘n’ times, then(i) Probability of occuring the event exactly ‘r’ times is nC

r P r (1 – P) n – r.

(ii) Probability of occuring the event atleast ‘r’ times is

n

rk

knkk

n )P1(PC .

(iii) Probability of occuring the event atmost ‘r’ times is

r

0k

knkk

n )P1(PC .

For example if a die is thrown five times and we want the probability of occurrence of a composite

number four times, then we have p =2 1 1 2

, q 16 3 3 3 and n = 5, r = 4. Thus required

probability = nCrprqn – r = 5C

4

4 11 2 103 3 243

.

Drill Exercise - 4

1. A policeman fires six bullets on a dacoit. The probability that the dacoit will be killed by one bullet is0.6, then find the probability that dacoit is still alive.

2. One hundred identical coins, each with probability p, of showing up heads are tossed. If0 < p < 1 and the probability of heads showing on 50 coins is equal to that of the heads showing in 51coins, then find the value of p.

3. If a pair of fair dice is rolled 5 times, then find out the probability that 3 times we get sum more than 9.

4. If 4 times two cards are drawn from a pack of playing cards (each time replacing both). Then find theprobability that only one time, we get a pair. (cards of same denomination)

5. A number is selected 3 times with replacement from the set {10, 11, 12 . . . 99}. Find out theprobability that at least two times product of digits of the selected number is 18.


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6. CONDITIONAL PROBABILITYThe probability of occurrence of an event B when it is known that some event A has already occurred,is called the conditional probability and is denoted by P(B/A). The symbol P(B/A) is usually read asprobability of B, given A.Consider two events A and B. When it is known that event ‘A’ has occurred,it means that sample space would reduce to that sample points representing event A. Now for P(B/A)we must look for the sample points representing the simultaneous occurrence of A and B i.e., samplepoints in A B .

P(B/A) =

n A Bn A B P(A B)n(S)

n(A)n A P(A)n(S)

Thus P(B/A) =

P A B

P A

,where 0 < P(A) 1. Similarly, P(A/B) =

P A B

P B

,0 < P(B) 1

Hence, P A B =P(A).P(B/ A), P(A) 0

P(B).P(A / B), P(B) 0

Illustration 4:If a pair of dice is thrown and it is known that sum of the numbers is even, then find the probability thatthe sum is less than 6.

Solution :Let A be the given event and let B be the event, whose probability is to be found. Then

Required probability P B AB 4 / 36 2

PA P(A) 18 / 36 9

.

Drill Exercise - 5

1. There is 30% chance that it rains on any particular day. What is the probability that there is at leastone rainy day within a period of 7-days ? Given that there is at least one rainy day, what is theprobability that there are at least two rainy days.

2. A fair coin is tossed repeatedly. If tail appears on first four tosses, then find the probability that headappearing on the fifth toss.

3. In a certain town, 40% of the people have brown hair, 25% have brown eyes and 15% have bothbrown hair and brown eyes. If a person selected at random from the town, having brown hair, thenfind the probability that he also has brown eye.

4. If E and F are the complementary events of events E and F respectively and if 0 < P(F) < 1, thenprove that

(i) P(E / F) + P E / F 1 (ii) P(E / F) E / F 1

5. A coin is tossed until a head appears or until it has been tossed 3 times. Given that head does notappear on the first toss, then find the probability that the coin is tossed three times.


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6.1 Independent EventsTwo events A and B are said to be independent if occurrence or non–occurrence of one does notaffect the occurrence or non–occurrence of the other,i.e., P(B/A) = P(B), P(A) 0.

P(B/A) = P A B

P(B)P(A)

P(A B) = P(A). P(B)

If the events are not independent, then they are said to be dependent.

Illustration 5 :If A and B are independent events. Then show that the following pairs of events are alsoindependent (i) A and B (ii) A and B

Solution :

(i) P A B P B P A B P B P(A)P(B)

( A and B are independent)

= P(B) (1 – P(A)) = P(B) P A .

(ii) P A B P A B 1 P A B

= 1 P(A) P(B) P(A).P(B)

= 1 – P(A) – P(B) + P(A). P(B) = (1 – P(A)) (1 – P(B)) = P A . P B .

6.2 Mutually independent EventsThree events A, B and C are said to be mutually independent if,

P A B = P(A). P(B), P A C = P(A). P(C), P B C = P(B). P(C)

and P A B C = P(A). P(B). P(C)

These events would be said to be pair–wise independent if,

P A B = P(A). P(B), P B C = P(B). P(C) and P A C = P(A). P(C).

Thus mutually independent events are pair–wise independent but the converse may not be true.

Illustration 6 :A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at atime with replacement. The events A, B and C are defined as under:A = {The first bulb is defective}, B = {The second bulb is non-defective}C = {The two bulbs are either both defective or both non-defective}Catogorize the events A, B and C to be pairwise independent or mutually independent.


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Solution :

P(A) =1 1

12 2 , P(B) =

1 11

2 2 and P(C) =

1 1 1 1 12 2 2 2 2

P A B P (the first bulb in defective and the second bulb is non-defective)

=1 1 12 2 4 .

P B C = P(both the bulbs are non-defective) =1 1 12 2 4

P C A = P (both the bulbs are defective) =1 1 12 2 4

P A B C P 0

As P A B P(A).P(B),P B C P B .P C and P C A P(C).P(A) , the events

A, B and C are pairwise independent.

Since P A B C 0 P(A).P(B).P(C) , A, B and C are not mutually independent.

6.3 Rule of Multiplication

Suppose A1, A

2, . . . , A

n be n events such that 1 2 nP A A . . . A 0 , then

n

i 1 2 1 3 1 2 n 1 2 n 1i 1

P A P A .P A / A .P A / A A . . .P A / A A . . . A

Illustration 7:P

1, P

2, . . . , P

8 an eight players participating in a tournament. If i < j, then P

i will win the match against

Pj. Players are paired up randomly for first round and winners of this round again paired up for the

second round and so on. Find the probability that P4 reaches in the final.

Solution:Let A

1 be the event that in the first round the four winners are P

1, P

4, P

i, P

j, where i {2, 3},

j {5, 6, 7}and let AA2 be the event that out of the four winners in the first round, P

1 and P

4 reaches

in the final.The event A

1 will occur, if P

4 plays with any of P

5, P

6, P

7 or P

8 (say with P

6) and P

1, P

2 and P

3 are

not paired with P5, P

7 and P

8. Further A

2 will occur if P

1 plays with P

j.

The required probability = P 1 2A A = P(A1). P

2

1

AA

.

=

3

4 2

64 3

2 3 18 4

2 4 2 2

= 4

3 5.

(Here we have used the concept of division into groups).


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drill exercise - 6

1. Find the least number of times must a fair die be tossed, in order to have a probability of at least91

216,

of getting atleast one six.

2. Two subsets A and B of the set {1, 2, . . ., n} are chosen at random. Find the probability that they donot have any element in common.

3. Two cards are drawn at random from a pack of playing cards. Find the probability that one cardis a heart and the other is an ace.

4. Let ‘p’ is the probability that a man aged 75 years will die in a year, then find the probability thatamong the persons S

1, S

2, S

3,.........S

50 each aged 75 years, S

1 will die in a year and will be the first

one to die.

5. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards, then findthe probability of drawing two aces.

6.4 Total Probability TheoremSuppose A

1, A

2 . . . , A

n are mutually exclusive and exhaustive events, then for any event B, we can

write B = n

i 1

B A

n

ii 1

P B P B A

, as events iB A , i = 1, 2, . . ., n are exclusive.

P(B) = n

i ii 1

P B / A .P A .

Illustration 8 :Two sets of candidates are competing for the positions on the board of directors of a company. Theprobabilities that the first and second sets will win are 0.6 and 0.4 respectively. If the first set wins, theprobability of introducing a new product is 0.8, and the corresponding probability, if the second setwins is 0.3. What is the probability that the new product will be introduced?

Solution :Let A

1 (A

2) denotes the event that first (second) set wins and let B be the event that a new product

is introduced. P(A

1) = 0.6, P(A

2) = 0.4

P1 2

B B0.8, P

A A

= 0.3

P(B) = 1 2 1 21 2

B BP B A P B A P A .P P A P

A A

.

= 0.6 0.8 0.4 0.3 0.6 .


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6.5 Baye’s TheoremSuppose A

1, A

2, . . ., A

n are mutually exclusive and exhaustive set of events.

Then the conditional probability that Ai happens (given that B has happened) is given by

P(Aj/B) =

j j j

n

i ii 1

P A B P A .P B / A

P BP A .P B / A

, where j = 1, 2, ....., n.

Illustration 9 :Box I contains 2 white and 3 red balls and box II contains 4 white and 5 red balls. One ball isdrawn at random from one of the boxes and is found to be red. Find the probability that it wasfrom box.

Solution :Let A denote the event that the drawn ball is redLet A

1 The event that box I is selected and let AA2 The event that box II is selected

222

1 21 2

AP A .P

AAP

A A AP A P P A .P

A A

=

1 5. 252 9

1 5 1 3 32. .2 9 2 5

Drill Exercise - 7

1. Three groups A, B and C are competing for position on the Board of Directors of a company. Theprobabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability ofintroducing a new product is 0.7 and the corresponding probabilities for groups B and C are 0.6and 0.5 respectively. Find the probability that the new product will be introduced.

2. A factory A produces 10% defective values and another factory B produces 20% defective. A bagcontains 4 values of factory A and 5 valves of factory B. If two valves are drawn at random fromthe bag, find the probability that atleast one value is defective. Give your answer upto two placesof decimals.

3. A man is known to speak truth 3 out of 4 times. He throws a dice and reports that it is six, then findthe probability that it is actually six.

4. A bag contains 3 red and 4 white balls. If a fair coin is tossed, if head comes two ball is drawn and iftail comes three ball is drawn, then find the probability that draw contains balls of both the colours.

5. A bag contains 10 fair coins and 25 coins having heads on both sides. A coin is selected at randomlyand tossed. If it gives head, then find out the probability that it was a fair coin.


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Answer Key

Drill Exercise - 1

1. 0.55 2. 5/216 3. 0.31

4. 1/17 5.15

7

Drill Exercise - 2

1. 319/420 2.3

1,

9

13.

209

243

4. 53/80 5. 1/3, 1/12

Drill Exercise - 3

1. (i) 0.82 (ii) 0.76

2. (i)34

(ii)12

(iii)7

12(iv)

58

(v)56

3. 0.25 4. 35.)CB(P2. 5.2

1 and

3

1 or

3

1 and

2

1

Drill Exercise - 4

1. 0.004096 2.51

1013. 56

2504. 4

14

17

25. 345

524

Drill Exercise - 5

1. [1 - (7/10)7 - 7C1(3/10) (7/10)6 / 1 - (7/10)7] 2.

2

13.

38

5.1

2

Drill Exercise - 6

1. 3 2.n

3

4

3.26

14.

1 1 ( )pn

n

5.13

1

13

1

Drill Exercise - 7

1. 0.63 2. 517/1800 3.4

3

4.7

55.

6

1


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SOLVED SUBJECTIVE EXAMPLES

Example 1:If m different cards are placed at random and independently into n boxes lying in a straight line(n > m), find the probability that the cards go into m adjacent boxes.

Solution :m cards can be placed into n boxes independently in nm ways. n(S) = nm

Now m adjacent boxes can be chosen in n – m + 1 ways. In each of these the cards can be placedinto m! ways.total number of ways in which the cards can be placed in m consecutive boxes

= (n – m + 1). m!

Required probability =

m

m! n m 1

n

.

Example 2 :Out of 21 tickets consecutively numbered, three are drawn at random. Find the probability that thenumbers on them are in A.P.

Solution :Any three tickets out of 21 tickets can be chosen is 21C

3 ways. For the favourable choice if the chosen

numbers are a, b and c, a < b < c, then we should havea c

b2

. Obviously either both a and c are

even or both are odd and then b is fixed. Hence for the favourable choice we have to chose twonumbers from 1 to 21, which are either both even or both odd. This can be done in11C

2 + 10C

2 ways.

Hence required probability =11 10

2 221

3

C C 10

133C

.

Example 3 :A has 3 shares in a lottery containing 3 prizes and 9 blanks. B has 2 shares in a lottery containing 2prizes and 6 blanks. Compare their chances of success.

Solution :Let E

1 be the event of success of A and let E

2 be the event of success of B

Since A has 3 shares in a lottery containing 3 prizes and 9-blanks, A will draw 3 tickets out of 12tickets (containing 3 prizes and 9 blanks). A will get success if he draws atleast one prize out of 3draws.

9

31 12

3

C 21P E

55C

121 34

P E 155 55

Again, 6

22 8

2

C 15P E

28C


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Probability

P(E2) = 1 – 15 13

28 28

1

2

P E 34 28 95255 13 715P E

P(E1) : P(E

2) = 952 : 715.

Example 4:A coin is tossed m + n times (m > n). Show that the probability of at least m consecutive heads come

up is m 2

n 2

2

.

Solution :Let H, T and S be the events “head turns up”, “tail turns up” and “head or tail turns up”

Then P(H) = P(T) =1

2 and P(S) = 1

Since the given event is “at least m consecutive heads turn up”, therefore in any favorable out comethere are m consecutive heads and the rest are any of head or tailConsider the events

A1 =

m times n times

H,H,H,. . .,H ,S,S,S,. . .,S with P(A

1) =

nm m

1 1.1

2 2

A2 =

m times n 1times

T,H,H,H,. . .,H ,S,S,S,. . .,S

with P(A

2) =

n 1m m 1

1 1 1. .1

2 2 2

A3 =

m times n 2 times

S,T,H,H,H,. . .,H ,S,S,S,. . .,S

with P(A3) = 1.

n 2m m 1

1 1 1. .1

2 2 2

. . . An + 1

=n 1 times m times

S,S,S,. . .,S ,T,H,H,H,. . .,H

With P(A

n + 1) = 1n – 1. m m 1

1 1 1.

2 2 2

The given event is 1 2 3 n 1A A A A . As AA1, A

2, A

3, . . ., A

n + 1 are pair – wise mutually exclusive.

The required probability

= P(A1) + P(A

2) + P(A

3) + . . . + P(A

n + 1) = m m 1 m 1 m 1

n times

1 1 1 1. . .

2 2 2 2

=m m 1 m 1

1 n 2 n

2 2 2

.

Example 5:There are four six faced dice such that each of two dice bears the numbers 0, 1, 2, 3, 4 and 5 and theother two dice are ordinary dice bearing numbers 1, 2, 3, 4, 5 and 6. If all the four dice are thrown,find the probability that the total of numbers coming up on all the dice is 10.

Solution :Total number of sample points in the sample space = 64 = 1296Number of sample points in favour of the event


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Probability

= Coefficient of x10 in the expansion of (1 + x + x2 + . . . + x5)2 (x + x2 + . . . + x6)2

= Coefficient of x10 in the expansion of x2(1 + x + x2 + . . . + x5)4

= Coefficient of x8 in the expansion of (1 + x + x2 + . . . + x5)4

= Coefficient of x8 in the expansionof

461 x

1 x

= Coefficient of x8 in the expansion of (1 – x6)4 (1 – x)–4

= Coefficient of x8 in the expansion of (1 – 4x6)2 34 5 4 5 6

1 4x x x . . .2! 3!

= 1 11 58 2C 4 C 125 .

Required probability =125

1296.

Example 6:A die is thrown 7 times. What is the probability that an odd number turns up (i) exactly 4 times (ii)atleast 4 times.

Solution :

Probability of success =3 1

6 2 p =

1

2 , q =

12

(i) For exactly four successes, required probability = 7C4.

4 31 1 35

.2 2 128

(ii) For atleast four successes, required probability

= 7C4

4 3 5 2 6 1 77 7 7

5 6 71 1 1 1 1 1 1

. C . C . C2 2 2 2 2 2 2

=35 21 7 1 64 1

128 128 128 128 128 2 .

Example 7:If m things are distributed among ‘a’ men and ‘b’ women, show that the probability that the number of

things received by men is odd, is

m m

m

b a b a12 b a

.

Solution :

A particular thing is received by a man with probability p =a

a band by a woman with probability q

=b

a b. If distributing a single object is an experiment, then this experiment is repeated m time. The

required probability = m1C . p. qm – 1 + mC

3 . p3. qm – 3 + mC

5 . p5. qm – 5 + . . .


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Probability

= m mq p q p

2

=

m1 b a

12 b a

=

m m

m

b a b a12 b a

.

Example 8 :

An artillery target may be either at point A with probability89

or at point B with probability19

. WeWe

have 21 shells each of which can be fixed either at point A or B. Each shell may hit the target

independently of the other shell with probability12

. How many shells must be fired at point A to hit the

target with maximum probability?

Solution :Let E denote the event that the target is hit when x shells are fired at point A. Let E

1 (E

2) denote the

event that the artillery target is at point A B .

We have P(E1) =

89

, P(E2) =

19

.

Px 21 x

1 2

E 1 E 11 and P 1

E 2 E 2

Now P(E) =89

x 21 x1 1 11 1

2 9 2

x 21 xd 8 1 1 1

P(E) n 2 n2dx 9 2 9 2

Now we must have dP E 0

dx

x = 12, also 2

2

dP E 0

dx

Hence P(E) is maximum, when x = 12.

Example 9:Let p be the probability that a man aged x years will die within a year. Let A

1, A

2, . . . , A

n be n men

each aged x years. Find the probability that out of these n men A1 will die with in a year and is first to

die.Solution :

P(no one among A1, A

2 . . . , A

n dies within a year) = (1 – p)n

P (at least one among A1, A

2, . . ., A

n dies within a year) = 1 – (1 – p)n

P(A1dies within a year and is first to die) = n1

1 1 pn .


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Example 10:Each of three bags A, B, C contains white balls and black balls. A has a

1 white & b

1 black, B has a

2

white & b2 black and C has a

3 white & b

3 black balls. A ball is drawn from a bag and found to be

white. What are the probabilities that the ball is from bag A, B and C.

Solution :Let A

1, A

2 and A

3 be the events that the bag picked is A, B and C respectively.

Let E be the event that a white ball is drawn.We are supposed to find P(A

1/E), P(A

2/E), P(A

3/E).

P1A

E

= 1P A E Probability that bag Ais chosen and white ball is drawn

P(E) Probability that a bag is chosen at random and a white ball is drawn

=

1 1

1 1 2 2 3 3

P A .P E / A

P A .P E / A P A .P E / A P A .P E / A

=

1

11 1

1 2 31 2 3

1 1 2 2 3 3

1 a.

p3 a bp p p1 a a a

.3 a b a b a b

, pk

= k

k k

a

a b, k = 1, 2, 3.

Similarly, p(A2/E) = 2

1 2 3

p

p p p , p(A

3/E) = 3

1 2 3

p

p p p .

Example 11:The probability that at least one of A and B occurs is 0.6. If A and B occur simultaneously with

probability 0.3, then find the value of )B(P)A(P .Solution :

We have 6.0)BA(P and 3.0)BA(P

P(A) + P(B) = )BA(P + )BA(P = 0.6 + 0.3 = 0.9

1.19.02)B(P1)A(P1)B(P)A(P

Example 12:There are n students in a class and probability that exactly out of n pass the examination is directly

proportional to 2 0 n .

(i) Find out the probability that a student selected at random was passed the examination.(ii) If a selected student has been found to pass the examination then find out the probability thathe is the only student to have passed the examination.

Solution :Let E be the event that exactly out of n pass the examinations and let A be the event that a student

selected randomly pass the examination.

2P E


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Probability

2P E k (k is proportionality constant)

Since E0, E

1, E

2, . . . , E

n are mutually exclusive and exhaustive events.

P(E0) + P(E

1) + P(E

2) + . . . + P(E

n) = 1

0 + k (1)2 + k (2)2 + . . . + k (n2) = 1

k = 6

n n 1 2n 1 . . . (i)

(i) n

0

P(A) P E P A / E =

n n2 3

1 1

3 n 1kk

n n 2 2n 1

(ii)

2111

AP E .P

EE 2P

A P A n n 1

P A EAP

nP E

.

Example 13 :Let A and B be two independent witnesses in a case. The probability that A will speak the truth is xand the probability that B will speak the truth is y. A and B agree in a certain statement. Show that the

probability that the statement is true isxy

1 x y 2xy .

Solution :Let E

1 be the event that both A and B speak the truth, E

2 be the event that both A and B tell a lie and

E be the event that A and B agree in a certain statement. Let C be the event that A speaks the truth andD be the event that B speaks the truth.

E1 = C D and E

2 = C D . P(E

1) = P C D = P(C) P(D) = xy and

P(E2) = P C D = P C P D = (1 – x) (1 – y) = 1 – x – y + xy

Now P1

EE

= probability that A and B will agree when both of them speak the truth = 1 and

P2

EE

= probability that A and B will agree when both of them tell a lie = 1.

Clearly,1E

E

be the event that the statement is true.

P

1 11

1 1 2 2

P E .P E / EEE P E .P E / E P E P E / E

= xy.1 xy

1 x y 2xyxy.1 1 x y xy .1

.

Example 14 :


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Find the minimum number of tosses of a pair of dice, so that the probability of getting the sum of thenumbers on the dice equal to 7 on atleast one toss, is greater than 0.95.(Given log

102 = 0.3010, log

103= 0.4771).

Solution :n(S) = 36Let E be the event of getting the sum of digits on the dice equal to 7, then n(E) = 6.

P(E) =6 1

p36 6 , then P E = q =

56

probability of not throwing the sum 7 in first m trails = qm .

P(at least one 7 in m throws) = 1 – qm = 1 –m5

6

.

According to the questionm5

1 0.956

m5

0.056

10 10 10 10m log 5 log 6 log 1 log 20

m > 16.44Hence, the least number of trails = 17.

Example 15 :Two teams A and B play a tournment. The first one to win (n + 1) games, win the series. Theprobability that A wins a game is p and that B wins a game is q (no ties). Find the probability that A

wins the series. Hence or otherwise prove thatn

n rr n r

r 0

1C . 1

2

.

Solution :A wins the series, if out of first n + r games A wins n games, 0 r n and wins the (n + r + 1)th game.

P(A) =n

n r r nn

r 0

C .q .p

. p (where p + q = 1)

Similarly, P(B) = n

n r n 1 rn

r 0

C .q .p

Now P(A) + P(B) = 1

n

r n 1 n 1 r n rn

r 0

q .p q .p C 1

. . . (i)

Now put p = q =12

from (i), n

n rn n r

r 0

1C . 1

2

.


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SOLVED OBJECTIVE EXAMPLES

Example 1 :A natural number is chosen at random from the first one hundred natural numbers. The probability that

x 20 x 400

x 30

is

(a)1

50(b)

350

(c)325

(d)725

Solution :From the wavy curve method, given inequality is satisfied for x < 20 or 30 < x < 40.

Number of favourable outcomes = 28

Required probability =28 7

100 25 .

Example 2 :

If A and B are two events such that 5P(A B)

6 , P 1 3

A , p(B)3 4 , then A and B are

(a) mutually exclusive (b) dependent(c) independent (d) none of these

Solution :

P A B P(A) P(B) P A B

1 3 5 3 1P A B

3 4 6 12 4

P(A) P(B) =1 3 1

.3 4 4 .

As P(A) P(B) = P A B independent

P A B 0 not exclusive.

Example 3 :

If1 3p 1 p

,3 2

and

1 p

2

are the probabilities of three mutually exclusive events, then the set of all

values of p is

(a) (b)1 1

,2 3

(c) [0, 1] (d) none of these


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Solution :We have

0 1 3p 1 p

,3 2

and

1 2p 1 11 p ,

2 3 2

. Further if the events

(say E1, E

2 and E

3) are exclusive, then its necessary and sufficient condition is

P 1 2 3 1 2 3 1 2 38 3p

E E E P E P E P E P E E E6

0 8 3p 2 8

1 p ,6 3 3

.

Hence the required set is .

Example 4 :

For independent events A1, . . ., A

n, P(A

i) =

1,

i 1 i = 1, 2, . . ., n. Then the probability that none

of the events will occur is(a) n/(n + 1) (b) n – 1/(n + 1)(c) 1/(n + 1) (d) none of these

Solution :P(non occurrence of (A

i)) = 1 – (i + 1) = i/(i + 1).

P(non occurrence of any of events)

= 1 2 n 1

. . . .2 3 n 1 n 1

.

Example 5 :A bag contains a large number of white and black marbles in equal proportions. Two samples of 5marbles are selected (with replacement) at random. The probability that the first sample containsexactly 1 black marble, and the second sample contains exactly 3 black marbles, is

(a)25

512(b)

15

32

(c)15

1024(d)

35256

Solution :Let the number of marble be 2n (where n is large)

Required probability =n nn

3 242n 2nn 5 5

C Cn Clim

C C

=

22

2n

5 2n 5 !n n n 1 n 2 n 3 n n 1 n 2 n n 1lim

4! 3! 2! 2n!


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Probability

=

23 24

2n

n n 1 n 2 n 3 2n 5 ! 5 5 4 3!lim

3!2! 2n!

=

3 24

2n

50.n n 1 n 2 n 3 50 25lim

1024 5122n 2n 1 2n 2 2n 3 2n 4

.

Example 6 :If two events A and B are such that P A = 0.3, P(B) = 0.4 and A B = 0.5, then

PB

A B

=

(a)14

(b)15

(c)35

(d)2

5Solution :

P B/ A B =

P B A B

P A B

= P A B P(A) P(A B )

P(A) P(B ) P(A B ) 0.7 0.6 0.5

=0.7 0.5 1

0.8 4

.

Example 7 :A is a set containing n elements. A subset P

1 of A is chosen at random. The set A is reconstructed by

replacing the elements of P1. A subset P

2 is again chosen at random. The probability that 1 2P P contains

exactly one element, is

(a) n

3n

4(b)

n

n

3

4

(c)3

4(d) none of these

Solution :Any element of A has four possibilities : element belongs to (i) both P

1 and P

2

(ii) neither P1 nor P

2 (iii) P

1 but not to P

2 (iv) P

2 but not to P

1. Thus n(S) = 4n. For the favourable

cases, we choose one element in n ways and this element has three choices as (i), (iii) and (iv), whilethe remaining n – 1 elements have one choice each, namely (ii).

Hence required probability =n

3n

4.

Example 8 :The probability that in a group of N (< 365)people, at least two will have the same birthday is


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(a)

365 !

1365 N ! 365 !

(b)

N365 365 !

365 N ! – 1

(c) 1 –

N365 365 !

365 N ! (d) none of these

Solution :Let A be the event of different birthdays. Each can have birthday in 365 ways, so N persons canhave their birthdays in 365N ways. Number of ways in which all have different birthdays = 365P

N

365NN N

365 !PP(A) 1 P A 1 1

365 365 365 N !

.

Example 9 :Let E and F be two independent events such that P(E) > P(F). The probability that both E and F

happen is1

12 and the probability that neither E nor F happens is

1

2, then

(a) P(E) =1

3, P(F) =

1

4(b) P(E) =

1

2, P(F) =

16

(c) P(E) = 1, P(F) =112

(d) none of these

Solution :

P E F = P(E) P(F) =1

12. . . (i)

P c c c c 1E F P E .P F

2

11 P E 1 P F

2 . . . (ii)

Solving (i) and (ii), we get P(E) =1 1

& P(F)2 4

, as P(E) > P(F).

Example 10:A draw two cards at random from a pack of 52 cards. After returning them to the pack andshuffling it, B draws two cards at random. The probability that there is exactly one common card,is

(a)5

546

(b)

50

663

(c)25

663(d) none of these

Solution :


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Let S be the sample space and let E be the required event, then n(S) = 2522C . For the number of

elements in E, we first choose a card (which we want common) and then from the remaining cards (51in numbers) we choose two cards and distribute them among A and B in 2! ways. Hence n (E) = 52C

1.

51C2. 2!. Thus P(E) =

50

663.

Example 11: :A company has two plants to manufacture televisions. Plant I manufacture 70% of televisions andplant II manufacture 30%. At plant I, 80% of the televisions are rated as of standard quality and atplant II, 90% of the televisions are rated as of standard quality. A television is chosen at randomand is found to be of standard quality. The probability that it has come from plant II is

(a)1750

(b)2783

(c)35

(d) none of these

Solution :Let E be the event that a television chosen randomly is of standard quality. We have to find

P(II/E) =

P E / II .P(II)

P E / I .P I P E / II .P II

=

9 /10 3/10

27 / 834 / 5 7 /10 9 /10 3/10

Example 12:x

1, x

2, x

3, . . . , x

50 are fifty real numbers such that x

r < x

r + 1 for r = 1, 2, 3, . . ., 49. Five numbers out

of these are picked up at random. The probability that the five numbers have x20

as the middle numberis

(a)20 30

2 250

5

C C

C

(b)

30 192 250

5

C C

C

(c)19 31

2 350

5

C C

C

(d) none of these

Solution :n(S) = 50C

5 = Total number of ways

n(E) = 30C2 × 19C

2 = Number of favourable ways

P(E) =30 19

2 250

5

C C

C

.

Example 13 :If the integers m and n are chosen at random from 1 to 100, then the probability that a number ofthe form 7n + 7m is divisible by 5 equals


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(a)1

4(b)

1

2

(c)1

8(d) none of these

Solution :We observe that 71, 72, 73 and 74 ends in 7, 9, 3 and 1 respectively. Thus 7l ends in 7, 9, 3 or 1according as l is of the form 4k + 1, 4k + 2, 4k – 1 or 4k respectively. If S is the sample space, thenn (S) = (100)2. 7m + 7n is divisible by 5 if (i) m is of the form 4k + 1 and n is of the form4k – 1 or (ii) m is of the form 4k + 2 and n is of the form 4k or (iii) m is of the form 4k – 1 and n isof the form 4k + 1 or (iv) m is of the form 4k and n is of the form 4k + 1.

Thus number of favourable ordered pairs (m, n) = 4 25 25 . Hence required probability is14

.

Example 14 :

The probability that a man can hit a target is3

4. He tries 5 times. The probability that he will hit the

target at least three times is

(a)291

364(b)

371

461

(c)471

502(d)

459

512Solution :

P =3 1

,q , n 54 4

Required probability = 5C3

3 2 4 55 5

4 53 1 3 1 3

C . C4 4 4 4 4

=459

512.

Example 15 :A die is thrown 7 times. The chance that an odd number turns up at least 4 times, is

(a)14

(b)12

(c)18

(d) none of these

Solution :For at least 4 successes, required probability

= 7C4

4 3 5 2 6 1 77 7 7

5 6 71 1 1 1 1 1 1

C C C2 2 2 2 2 2 2

=

12

.


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StudySteps - WordPress.com · used to describe the outcomes of an experiment. In this case ‘S 1 ’ provides more information than ‘S 2 ’. If we know which element in S 1 occurs,