Theoretical Distributions In Business Statistics 15 October

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Find the binomial distribution if the mean is 48 standard deviation is 4.


P(A) =1/3, P(B) = ½, P(A-B) = 1/5, what is the probability that only B

will happen?

• P(A) = 1/3 (WHICH INCLUDES A+B ALSO).

• P(A-B) = 1/5 (ONLY A – BECAUSE HERE B IS EXCLUDED)

• P(A &B) (Together) = 1/3-1/5 = 2/15• P(ONLY B) = ½ - 2/15 = 11/30 ans.


What is the probability that 4 children selected at random would

have different birthdays?

• For the first boy there are 365 choices, for second, there are 364 choices, for third there are 363 choices and for 4th there are 362 choices. Thus the probability is

• (365*364*363*362) / (365^4) = .983644 answer


There are 3 boxes with following details: Box 1. (5 red,7 white, 6

blue) box 2: (4red,8 white,6 blue), box III : (3 red,4 white,2 blue), if one ball is drawn at random from

each of these, what is the probability that we have 3 balls of

same colour?


Solution.

• Let us assume that we get Red from each: • 5/18*4/18* 3/9 = 60/2916• For white : • 7/18*8/18*4/9 = 224 /2916• For blue: • 6/18*6/18*2/9 = 72/2916• = 356/2916 : = .122 ans.


What is the probability that a leap year selected at random will have

53 Saturdays?

• We have 52 week plus one extra day in a normal year. This extra day can be any one from Mon. to Sun. In leap year, we have 2 extra days thus we can have these 2 extra days on any of MON+Tue or any combination of 2 days.

• Thus we have 2/7 probability that the leap year will have 53 Saturday. Ans.


A pocket of 10 components include 3 defectives. If 4 components are

selected from the pocket at random, what is the expected value

of the number of defectives? 0 1 2 3

7c4/10c4

7c3*3c1/10c4

7c2*3c2/10c4

7c1*3c3/10c4

.167 .5 .3 .0330 .5 .6 .099 1.199


30% failed in Acct. 40% failed in Eco and 50% failed in at least one

of these papers. If we randomly pick up a person, who is failed in Eco,what is the probability that he

has passed Acct? • Let us find the % for those who have failed

in both the papers. • 30+40 – X = 50• X = 20 thus there is 50% probability that

he has passed Accts.


A problem was given to Mayank, Goti and Nidhi (pgpse students),

the probability they will solve is 1/3, 1/5 and ½, what is the probability that the problem will be solved?

• Let us assume that Mayank doesn’t solve it: 2/3, Goti doesn’t solve it 4/5 and Nidhi also doesn’t solve it ½. All 3 don’t solve it :

• 2/3*4/5*1/2 = 8/30. thus problem will be solved : 22/30 or 11/15 answer.


If NC2 = NC5, what is N ?

• N = 2+5 = 7• Thus N = 7 ans.


How many numbers of 5 digits can be formed with 0,1,2,3,4,6,8?

• For the first digit we have 6 options, for 2nd digit we have 6 options (the digit used earlier will not be used), for 3rd digit we have 5 options, and for 4th digit we have 4 options and for last 5th digit we have 3 options). Multiply all these to get the answer:

• =6*6*5*4*3*• =2160 answer. • If repetition is allowed: • 6*7*7*7*7• =14406 ans.


There are 2 boxes. The first contains 3 red and 5 white balls.

The 2nd contains 4 red and 6 white ball. One ball is taken from first box and transferred to 2nd box. And then

another ball is drawn from the second box. What is the probability

that the second ball is Red?

• Let us assume that we draw red ball from first box and then we have 5 red in 2nd box


Solution ….• 3/8 *5/11 = 15/88• Suppose we draw white ball from first box: • 5/8*4/11 = 20/88 (multiply because there is &

between them). • Now add these two: • 35/88 answer. (add because there is OR

between them). • (in first possibility, we have red ball from the first

box and again red ball from 2nd box. In 2nd possibility, we have white from 1st and red from 2nd box.).


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Theoretical Distributions In Business Statistics 15 October