# Answer to Question #21942 in Quantitative Methods for rushil

Question #21942

there are 15 students in a class. assuming that each student is equally likely to have been born on any day of the week. find the probability that three or fewer were born on a monday.

Expert's answer

The probability of a student to be born at some day in a week is p=1/7.

Let X be the number of students from the class born on a Monday.

Then X has binomial distribution with parameters

n=15, p=1/7

Put

q=1-p =6/7

Then the probability P(X=k) is computed by the following formula:

P(X=k) = C^n_k p^k q^(n-k),

where

C^n_k = n! / (k!*(n-k)! )

We should find

P(X<=3) =P(X=0) + P(X=1) + P(X=2) + P(X=3).

Let us compute these numbers

P(X=0) = C^15_0 p^0 q^15 = (6/7)^15 = 0.099037

P(X=1) = C^15_1 p^1 q^14 = 15 * (1/7)* (6/7)^14 = 0.24759

P(X=2) = C^15_2 p^2 q^13 = (15*14/2)* (1/7)^2 * (6/7)^13 = 0.28886

P(X=3) = C^15_3 p^3 q^12 = (15*14*13/(3*2)) * (1/7)^3 * (6/7)^12 = 0.20862

Hence

P(X<=3) = 0.099037 + 0.24759 + 0.28886 + 0.20862 = 0.84411

Thus the probability that three or fewer were born on a Monday is 0.84411.

Let X be the number of students from the class born on a Monday.

Then X has binomial distribution with parameters

n=15, p=1/7

Put

q=1-p =6/7

Then the probability P(X=k) is computed by the following formula:

P(X=k) = C^n_k p^k q^(n-k),

where

C^n_k = n! / (k!*(n-k)! )

We should find

P(X<=3) =P(X=0) + P(X=1) + P(X=2) + P(X=3).

Let us compute these numbers

P(X=0) = C^15_0 p^0 q^15 = (6/7)^15 = 0.099037

P(X=1) = C^15_1 p^1 q^14 = 15 * (1/7)* (6/7)^14 = 0.24759

P(X=2) = C^15_2 p^2 q^13 = (15*14/2)* (1/7)^2 * (6/7)^13 = 0.28886

P(X=3) = C^15_3 p^3 q^12 = (15*14*13/(3*2)) * (1/7)^3 * (6/7)^12 = 0.20862

Hence

P(X<=3) = 0.099037 + 0.24759 + 0.28886 + 0.20862 = 0.84411

Thus the probability that three or fewer were born on a Monday is 0.84411.

## Comments

## Leave a comment