Question

A jury of 12 people is to be randomly selected from a group of 30 people (18 women and 12 men). Let the random variable X = “number of women on the jury of 12.”

Find (using R):

a) P(X = 6)

b) P(X ≤ 8)

c) P(5 ≤ X ≤ 10)

d) P(no men on the jury

Answer 1

I have done this problem in R. So, I am attaching my R-code as well as a screenshot of my solution:

> #A jury of 12 people is to be randomly selected from a group of 18 women and > #X is the number of women on the jury of 12. > # a) P(X-6) > # P (X 6) (18C6 * 12C6 )/30C6 > (choose (18, 6)choose (12, 6))/choose (30, 12) [1] 0.1983177 > # P (X<=8)= 1-P (X>8)= 1-P(x-9) + (X=10) + P (x-11) + P (X=12) > 1-(choose (18, 9)choose (12, 3))/choose (30,12) (choose (18,10)* choose (12,2))/s [1] 3 > # > # 0.9143526 c) P ( 5<=X<=10) P (X=51+ P (X=61+ P (X=7 ) +p (X=81+P (X=91+P (X=10) > (choose (18,5)* choose (12,7))/choose (30,12)(choose (18,6) choose (12, 6))/chs I11 0.975663 > # d) P (no men on the jury) > # P (X 12) > (choose (18, 12)choose (12,0))/choose (30,12) 1] 0.0002146295

#A jury of 12 people is to be randomly selected from a group of 18 women and 12 men.
#X is the number of women on the jury of 12.
# a)P(X=6)
# P(X=6)= (18C6 * 12C6 )/30C6
(choose(18,6)* choose(12,6))/choose(30,12)

# b) P(X<=8)
# P(X<=8)= 1-P(X>8)= 1-P(X=9) + (X=10) + P(X=11) + P(X=12)
1-(choose(18,9)* choose(12,3))/choose(30,12) + (choose(18,10)* choose(12,2))/choose(30,12) + (choose(18,11)* choose(12,1))/choose(30,12) + (choose(18,12)* choose(12,0))/choose(30,12)

# c) P(5<=X<=10)
# P(X=5)+ P(X=6)+ P(X=7)+P(X=8)+P(X=9)+P(X=10)
(choose(18,5)* choose(12,7))/choose(30,12) + (choose(18,6)* choose(12,6))/choose(30,12) + (choose(18,7)* choose(12,5))/choose(30,12) + (choose(18,8)* choose(12,4))/choose(30,12) + (choose(18,9)* choose(12,3))/choose(30,12) + (choose(18,10)* choose(12,2))/choose(30,12)

# d)P(no men on the jury)
# P(X=12)
(choose(18,12)* choose(12,0))/choose(30,12)