Question

My daughter’s class has 18 students. On Wednesday they will go on a field trip. The class will be divided randomly (all equally likely to be placed in each group) into two groups of nine students each. I will chaperone one of the two groups. There are five rowdy boys in the class who cause lots of trouble. Let X = the number of rowdy boys in my group.

Show that X fits the criteria for a hypergeometric distribution and give the pmf for X.

What is the probability that there are fewer than two rowdy boys in my group?

Answer 1

Answer:

Class has all out 18 understudies so population size is given below

N= 18

Number of raucous young men in gathering is 9 so number of accomplishment in population is k=9

Your gathering has 9 arbitrarily choosen kid so test size is given below

n = 9

Furthermore, let us assume that x demonstrates the quantity of boisterous young men in your gathering so by the hypergeometric dissemination the likelihood of getting x rowdy young men ( x victories) in your gathering is given as

P(X = x) = kCx * (N-k)C(n-x) / (NCn)

substitute known values in above formula

= (5Cx)(13C(9-x)) / (18C9)

where as x = 0,1,2,3,4,5

Now the probability that there are fewer than two rowdy boys in my group is given as

P(X < 2) = summation of (5Cx)(13C(9-x)) / (18C9) [summation from 0 to 1]

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

P(X < 2) = 0.1471