Probability Solved Examples

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Solved Examples

Example 1:

A pair of dice is thrown. Find the probability of obtaining a sum of 8 or getting an even number on both the dice.

Solution:

Let the events be defined as:

A : obtaining a sum of 8

B : getting an even number on both dice

We are required to find out the total Probability A and B, i.e.

P(AUB) = P(A) + P(B) - P(AB)

Now cases favourable to A are

(3, 5) (5, 3) (2, 6) (6, 2) (4, 4)

So, P(A) = 5/36

Cases favourable to B : (2, 2), (2, 4), (2, 6),

(4, 2), (4, 4), (4, 6),

(6, 2), (6, 4), (6, 6).

P(B) = 9/36

Now, (2, 6) (6, 2) and (4, 4) are common to both events A and B

So P(AB) = 3/36

=> P(AUB) = = 5/36+9/36-3/36=11/36 (Ans.)



Example 2:

A number x is selected from first 100 natural numbers. Find the probability that x satisfy the condition x + 100/x > 50.

Solution:

Total number of ways of selecting x is 100.

Now the given condition is x + 100/x > 50, on analyzing this equation, carefully, we see that this equation is satisfied for all the numbers x such that x > 48 and also for x = 1 and 2

So, favourable number cases is 55

Therefore, probability = 55/100 = 11/20

Example 3:

'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. Show that A's chance of winning a prize of B's chance of winning a prizes in ratio 16:7

Solution:

Method 1:

'A' may win one, two or all the three prizes, the total probability is

=(3C16C2)/(9C3)+(3C26C1)/(9C3)+(3C36C0)/(9C3)=(45+18+1)/84=64/84=16/21

P(B) = ( 1C1)/( 3C1 )=1/3

(P(A))/(P(B))=(163)/21=16/7=> P(A) : P(B) = 16 : 7 (Proved)

Method 2:

A can win one, two or all the prizes in the following manner

1 prize, 2 blanks 2 prizes, 1 blank 3 prizes

I II III



All the three cases where 'A' can win is independent and mutually exclusive. Hence probability that A can win prize is given as

P(A) = P(I) + P(II) + P(III)

Let's calculate P(I): it is something like this:

A bag contains 3 white balls (prizes and 6 black balls (blanks) then what is the probability that out of 3 draws random, there is 1 white ball (prize) and 2 black balls (blanks).

Which one can calculate very easily.

Favourable ways for this event = 3C1 6C2

Total number of ways = 9C3

P(I) = ( 3C1 6C2)/(9C3),, similarly we can calculate P(II) and P(III)

Hence, P(A) = (3C1 6 C2)/(9C3)+(3C2 6C1)/(9C3)+( 3C3 6C0)/(9C3) = 16/21

B has only one share in a lottery, which consists 1 prize and 2 blanks' therefore B can at most win only one prize. Hence

P(B) = 1C1/3C1 = 1/3

(P(A))/(P(B))=(163)/21=16/7

P(A) : P(B) = 16:7 (Ans.)

Example 4:

There are three events A, B and C one of which must, and only one can happen, the odd are 8 to 3 against and 2 to 5 for B. Find the odd, against C.

Solution:

P(A) = 3/11, P(B) = 2/7, P(C) = x (say)

Since one most and only one can happen therefore A, B, C are mutually exclusive and exhaustive events.



So, P(A) + P(B) + P(C) = 1

=> 3/11+2/7+x=1

x = (77-21-22)/77=34/77

Odd against C = (77 - 34) : 34 or 43 : 34 (Ans.)

Example 5:

An unbiased coin is tossed. If the result is head, a pair of unbiased dice are rolled and the number obtained by adding the number on the two faces are noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4......, 12 is picked & the number on the card is noted. What is the probability that the number noted is 7 or 8.

Solution:

Let us define the events

A : head appears.

B : Tail appears

C : 7 or 8 is noted.

We have to find the probability of C i.e. P(C)

P(C) = P(A) P (C/A) + P(B) P(C/B)

Now we calculate each of the constituents one by one

P(A) = probability of appearing head = 1/2

P(C/A) = Probability that event C takes place i.e. 7 or 8 being noted when head is already appeared. (If something is already happen then it becomes certain, i.e. now it is certain that head is appeared we have to certainly roll a pair of unbiased dice).

= 11/36 (since (6, 1) (1, 6) (5, 2) (2, 5) (3, 4) (4, 3) (6, 2) (2, 6) (3, 5) (5, 3) (4, 4) i.e. 11 favourable cases and of course 6 6 = 36 total number of cases)



Similarly, P(B) = 1/2

P(B/C) = 2/11 (Two favourable cases (7 and 8) and 11 total number of cases).

Hence,P(C)=1/211/36+1/22/11=193/792 (Ans.)

Example 6:

If the probability for A to win a game against B is 0.4. If A has an option of playing either a "best of 3 games" or a "best of 5 games" match against B, which option should A choose so that the probability of his winning the match is higher?

Solution:

Lets define the events

A : A wins a game against B => P(A) = 0.4

: A losses a game against B => P ( ) = 0.6

B : A wins match against B

Case I: When A plays a "Best of three games" mathc.

In this case A will have to win at least two of three games if he has to win the match.

This he can do by winning either 2 games or all of the 3 games in the manner.

A A + A A + A A + A A A

I II III IV

P(B) = 3C2 A2 ( ) + 3C3 (A)3

= 3 (0.4)2 (0.6) + (0.4)3 = 0.352

Case II: When A play a "Best of 5 games against" B. Then

P(B) = 5C3 A3 ( )2 + 5C4 (A)4 + 5C5 (A)5



= 10 (0.4)3 (0.6)2 + 5 (0.4)4 (0.6) + (0.4)5 = 0.23424

Therefore, A should choose Best of three games (Ans.)

Example 7:

Three squares are chosen at random on a chessboard. Find the probability that they lie in a diagonal line.

Solution:

Total number of cases is 64C3. It is clear from the figure that favourable numbers of case are 2(3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 8C3. Now, diagonal can also be in the direction of NE to SW i.e. favourable cases are 2|2 (3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 8C3_|.

So required probability = (favourable cases)/(Total cases)

=(2|2(3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 8C3_|)/64C3 = 7/744 (Ans.) Example 8:

A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of the subset of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements.

Solution:

In set P we can have no element i.e. , 1 element, 2 element, ...... upto n elements. If we have no element in P, we will leave by all the elements and number of set Q formed by those elements



will have no common element in common with P. Similarly, it there are r elements in P we are left with rest of (n - r) element to form Q, satisfying the condition that P and Q should be disjoint.

Hence, the total number of ways in which P and Q are disjoint can be given by

= n

r=0 nCr (2)(n-r)=(3)n

[Suppose you have a 5 element set {1, 2, 3, 4, 5} and you formed a subset of 2 elements, which you can do in 5C2 ways. Let's select 1 and 2, then we are left with 3, 4 and 5. Now we can form (2)3 = 8 i.e. {}, {3}, {4}, {5}, {3, 4}, {3, 5}, {4, 5}, {3, 4, 5} subsets which will have no common element with the previously chosen subset. So the total number of ways in which you can form two disjoint set is 5C0(2)5 + 5C1(2)4 + 5C2(2)3 + 5C3(2)2 + 5C4(2)1 + 5C5(2)0 = (3)5]

Total number of ways in which we can form P and Q

= nC0 (2)n + nC1(2)n + nC2(2)n +...+ nCn(2)n = (4)n

or simply total number of ways = Total number of ways of forming P total number of ways of forming Q.

= 2n 2n = (4)n.

So, required probability = (3/4)n. (Ans.)

Example 9:

In a test an examinee either guesses or copies or knows that answer to a multiple choice question which has 4 choices. The probability that he makes a guess is 1/3 and the probability that he copies is 1/6. The probability that his answer is correct, given the copied it, is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly.

Solution:

P(g) = probability of guessing = 1/3

P(c) = probability of copying = 1/6

P(k) = probability of knowing = 1 - 1/3 - 1/6 = 1/2



(Since the three-event g, c and k are mutually exclusive and exhaustive)

P(w) = probability that answer is correct

P(k/w)=(P(w/k).P(k))/(P(w/c)P(c)+P(w/k)P(k)+P(w/g)P(g)) (using Baye's theorem)

= (11/2)/((1/8,1/6)+(11/2)+(1/41/3) )=24/29 (Ans.)

Example 10:

A speak truth 3 out of 4 times. He reported that Mohan Bagan has won the match. Find the probability that his report was correct.

Solution:

Method 1: Let T : A speaks the truth

B : Mohan Bagan won the match

Given, P(T) =3/4 . . P( ) = 1 - 1/3 = 1/4

A match can be won, drawn or loosen

. . P(B/T) = 1/3 P(B/ ) = 2/3 .

Using Baye's theorem we get

P(T/B) = (P(T).P(B/T))/(P(T).P(B/T)+P( )P(B/ ))

= 3/41/3)/(3/41/3+1/42/3)=(1/4)/(5/12)=3/5

Method 2: Let, T : The man speak truth

A : Mohan Bagan won the match

B : He reported that Mohan Bagan has won.

P(A) = 1/3(the match may also end in a draw)

P(T) = 3/4



P(B) = P(A) P(T) + P( ) P( )

= 1/33/4+2/41/4=1/4+1/6=(3+2)/12=5/12

P(T/B)=(P(B/T).P(T))/(P(B))=(1/33/4)/(5/12)=3/5 (Ans.)

Example 11:

A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random one at a time with replacement. The event A, B, C are defined as:

A : the first bulb is defective

B : the second bulb is non defective

C : both the bulbs are defective or both are non-defective,

Determine whether

(i) A, B, C are pair wise independent.

(ii) A, B, c are mutually independent.

Solution:

Let probability of defective bulb is denoted by Pd and non-defective bulb is denoted by Pn.

So, Pd = Pn = 1/2

Let, D1 : First bulb is defective

D2 : Second bulb is defective

N1 : First bulb is non-defective

N2 : Second bulb is non-defective

Now event A is First bulb is defective and second bulb is defective or First bulb is defective and second bulb is non-defective.

i.e. A is D1D2, D1N2



B is D1 N2, N1 N2

and C is D1D2, N1N2

=> AB = D1N2

BC = N1N2

CA = D1D2

=> P(AB) = 1/4 P(BC) = P(CA)

also P(A) = 1/2*1/2 + 1/2*1/2 = 1/2 = P(B) = P(C)

=> P(AB) = P(A) P(B)

P(BC) = P(B) P(C)

P(CA) = P(C) P(A)

This condition is insufficient for A, B, C to be pair wise independent

Now, ABC = => P(ABC) P(A) P(B) P(C)

Since, there is no element, which is common to all A, B, C.

So, A, B, C are not mutually independent events. (Ans.)

Example 12:

A sample consists of integer 1, 2......... 2n. The probability of choosing the integer k is proportional to log k. Find the conditional probability of choosing the integer 2 given that an even integer is chosen.

Solution:

P(integer k is chosen)k = C log k (. it is proportional to log k), where C is some constant.

Now let even

A : 2 is chosen



B : An even integer is chosen, we have to find conditional probability of A given event B has happened i.e. P(A/B)

P(A/B) = (P(AB))/(P(B))=(P(A).P(B/A))/(P(A)P(B/A))

= (C.log2)/(C log2+C log4+C log6+...+ C log2n)

= (log2)/log(2n.(|n)) = (log2)/(n log2 + log(|n)) (Ans) Example 13:

Two players A and B toss a coin alternatively, with A beginning the game. The players who first throw a head is deemed to be the winter. B's coin is fair and A's is biased and has a probability p showing a head. Find the value of p so that the game is equi-probable to both the players.

Solution:

Player 'A' wins if he gets head in the first trial or in third [If B does not get head in his first trial] and so on.

P(A) = p + (1 - p) 1/2 p + (1 - p) 1/2 (1 - p) 1/2 p +...

This is an infinite G.P. of first term p and common ratio (1-p)/2

= p/(1-((1-p)/2)) = (2p)/(p+1)

According to given condition

P(A) = P(B)

=> 2p/(1+p) = 1-2p/(p+1)

=> (4 p)/(1+p) = 1 => p=1/3 (Ans.)

Example 14:

In a multiple choice question, there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only when if he ticks all the correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, find the probability that he will get marks in the question.



Solution:

Let the multiple choice question has four alternatives in the form

(a) _____________ (b) _____________

(c) _____________ (d) _____________

The possible answers are

a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd

Hence the total number of possible answers of such questions = 15.

If a student is given 3 chances to make guess, he will definitely choose 3 different answers. Hence number of favourable way = 3.

Therefore, Probability = 3/15 = 1/5 (Ans.)

Alternative Method:

If you are bit practiced you can nearly visualize that there are 4 elements (options) to form number of subsets (possible answers), which we can do in 24 - 1 = 15 ways. (1 is subtracted, as null set in this case means you are not choosing any answer at all).

Total number of favourable way = 3

Hence probability = 3/15 = 1/5 (Ans.)

Example 15:

Four cards are drawn at random from a well-shuffled pack of 52 cards. Find the probability that there is exactly one pair.

Solution:

Let H, C, D, S denotes heat, club, diamond and spade respectively.

We have to draw 4 cards at random, so that it consists of exactly one pair. A pair means two cards of same denomination. i.e. (5H, 5C) or (6C, 6D), so any one of such (i.e. having exactly



one pair) out of four-selected card should look like.

2H, 2S, 3C, 5D

Now we will find in how many ways, we can do this.

We have a pack of card like this.

1 H, 1 C, 1 D, 1 S ...... (1)

2 H, 2 C, 2 D, 2 S ...... (2)

3 H, 3 C, 3 D, 3 S ...... (3)

J H, J C, J D, J S ...... (11)

Q H, Q C, Q D, Q S ...... (12)

K H, K C, K D, K S ...... (13)

Firstly, in how many ways can we select a pair? There are 13 rows in above depiction of a pack of card. Each row consists of 4 cards of same denomination. If we select any of the row, which we can do in 13C1 way (say we have selected 3 row) and in that selected row if we select any of the 2 cards (say 3 H and 3 C) out of 4 which we can do in 4C2 ways, we will get total number of selecting a pair.

i.e. 13C1 4C2

Having selected pair we have to select two cards of two different denominations.

Now we are left with 50 cards but out of those 50 cards 2 cards (in our case 3 D and 3 S) are of no use as they will destroy our pair (they will make either triplet or quartet). Hence we are left with only 48 cards.

Since we have to select 2 more cards and if one says it can be done in 48C2 ways. It will be wrong. In our case, as those 48C2 combinations will have many pairs included in itself, and we need only one pair. So what we need is to select two cards carefully. We will do it one by one. We will select one card, which we can do in 48C1 ways (say we have selected 5S).



Number of ways in which we have selected 3 cards

= 13C1 4C2 48C1.

Now we are left with 47 cards but out of this 47 cards there are 3 cards (5H, 5C, 5D) are of no use as selection of any one of them will form a pair. Therefore, we are left with only 44 cards. We can select any of one card without violating any of our conditions in 44C1 ways.

Hence favourable numbers of ways = 13C1 4C2 48C1 44C1

Total numbers of ways = 52C4

Probability = (13C14C248C144C1) / ( 52C4 ) (Ans.)

Example 16:

Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. Find the probability that these three numbers are in A.P.

Solution:

We have 21 tickets. The total number of ways of selecting 3 out of 21 tickets will be = 21C3.

We will calculate the number of favourable ways in the following manner, suppose first we have selected 1. Now other two selection can be done like this

10 ways.

Similarly, if we select number 2 then,

9 ways.

For starting with 3 we have 9 ways.



For starting with 4 we have 8 ways

So, the favourable ways

= 10 + 9 + 9 + 8 + 8 + 7 + 7 + 6 + 6 + 5 + 5 + 4 + 4 + 3 + 3 + 2 + 2 + 1 + 1 = 100

Probability = 100/21C3. (Ans.)

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Probability Solved Examples