Question

please explain the solution of question, very appreciated. l really have no idea about the solution

A committee of 7 persons was selected randomly from 10 men and 12 women. Then a sub-committee of 3 persons was selected randomly from the 7 committee members.

(c)Given that the committee has at least one man, find the expected number of women on the subcommittee.

(c)LetXbethe number ofwomen on the sub-committee. Then where: Y Hyp(22,123) EY 3x12/22 -1.636364 //22 10〉(12) 0.004643963. EY-PC,JE(YIC) 1.636364-0.004643963 × 3 1-0.004643963 1.630

Answer 1

7 persons selected from 10 men and 12 women

P[ no men ] = 12C7 / 22C7 = 792 / 170544 = 0.0046

P[ at least one men ] = 1 - ( 12C7 / 22C7 ) = 1 - 0.0046 = 0.9954

P[ 1 men in committee ] = 12C6 * 10C1 / 22C7 = 9240 / 170544 = 0.0542

P[ 2 men in committee ] = 12C5 * 10C2 / 22C7 = 792*45 / 170544 = 0.2089

P[ 3 men in committee ] = 12C4 * 10C3 / 22C7 = 495*120 / 170544 = 0.3483

P[ 4 men in committee ] = 12C3 * 10C4 / 22C7 = 220*210 / 170544 = 0.2709

P[ 5 men in committee ] = 12C2 * 10C5 / 22C7 = 66*252 / 170544 = 0.0975

P[ 6 men in committee ] = 12C1 * 10C6 / 22C7 = 12*210 / 170544 = 0.0148

P[ 7 men in committee ] = 12C0 * 10C7 / 22C7 = 1*120 / 170544 = 0.0007

subcommittee has 3 members out of 7

P[ 0 women in sub committee ] = P[ 0 women in sub committee | 1 men in committee ]*P[ 1 men in committee ] + P[ 0 women in sub committee | 2 men in committee ]*P[ 2 men in committee ] +P[ 0 women in sub committee | 3 men in committee ]*P[ 3 men in committee ] +P[ 0 women in sub committee | 4 men in committee ]*P[ 4 men in committee ] +P[ 0 women in sub committee | 5 men in committee ]*P[ 5 men in committee ] +P[ 0 women in sub committee | 6 men in committee ]*P[ 6 men in committee ] +P[ 0 women in sub committee | 7 men in committee ]*P[ 7 men in committee ]

P[ 0 women in sub committee | 1 men in committee ] = 0 ( sub committee has to have 3 men, but committee only has 1 men ]

P[ 0 women in sub committee | 2 men in committee ] = 0

P[ 0 women in sub committee | 3 men in committee ] = 3C3*4C0 / 7C3 = 0.0286

P[ 0 women in sub committee | 4 men in committee ] = 4C3*3C0 / 7C3 = 0.1143

P[ 0 women in sub committee | 5 men in committee ] = 5C3*2C0 / 7C3 = 0.2857

P[ 0 women in sub committee | 6 men in committee ] = 6C3*1C0 / 7C3 = 0.5714

P[ 0 women in sub committee | 7 men in committee ] = 7C3 / 7C3 = 1

P[ 0 women in sub committee | at least one men in committee ] = 0*0.0542 + 0*0.2089 + 0.0286*0.3483 + 0.1143*0.2709 + 0.2857*0.0975 + 0.5714*0.0148 + 1*0.0007 = 0 + 0 + 0.0099 + 0.0309 + 0.0278 + 0.0084 + 0.0007 = 0.0777

P[ 1 women in sub committee ] = P[ 1 women in sub committee | 1 men in committee ]*P[ 1 men in committee ] + P[ 1 women in sub committee | 2 men in committee ]*P[ 2 men in committee ] +P[ 1 women in sub committee | 3 men in committee ]*P[ 3 men in committee ] +P[ 1 women in sub committee | 4 men in committee ]*P[ 4 men in committee ] +P[ 1 women in sub committee | 5 men in committee ]*P[ 5 men in committee ] +P[ 1 women in sub committee | 6 men in committee ]*P[ 6 men in committee ] +P[ 1 women in sub committee | 7 men in committee ]*P[ 7 men in committee ]

P[ 1 women in sub committee | 1 men in committee ] = 0 ( sub committee has to have 2 men, but committee only has 1 men ]

P[ 1 women in sub committee | 2 men in committee ] = 2C2*5C1 / 7C3 = 0.1428

P[ 1 women in sub committee | 3 men in committee ] = 3C2*4C1 / 7C3 = 0.3428

P[ 1 women in sub committee | 4 men in committee ] = 4C2*3C1 / 7C3 = 0.5143

P[ 1 women in sub committee | 5 men in committee ] = 5C2*2C1 / 7C3 = 0.5714

P[ 1 women in sub committee | 6 men in committee ] = 6C2*1C1 / 7C3 = 0.4286

P[ 1 women in sub committee | 7 men in committee ] = 0

P[ 1 women in sub committee | at least one men in committee ] = 0*0.0542 + 0.1428*0.2089 + 0.3428*0.3483 + 0.5143*0.2709 + 0.5714*0.0975 + 0.4286*0.0148 + 0*0.0007 = 0 + 0.0289 + 0.1194 + 0.1393 + 0.0557 + 0.0063 + 0 = 0.3496

P[ 2 women in sub committee ] = P[ 2 women in sub committee | 1 men in committee ]*P[ 1 men in committee ] + P[ 2 women in sub committee | 2 men in committee ]*P[ 2 men in committee ] +P[ 2 women in sub committee | 3 men in committee ]*P[ 3 men in committee ] +P[ 2 women in sub committee | 4 men in committee ]*P[ 4 men in committee ] +P[ 2 women in sub committee | 5 men in committee ]*P[ 5 men in committee ] +P[ 2 women in sub committee | 6 men in committee ]*P[ 6 men in committee ] +P[ 2 women in sub committee | 7 men in committee ]*P[ 7 men in committee ]

P[ 2 women in sub committee | 1 men in committee ] = 1C1*6C2 / 7C3 = 0.4286

P[ 2 women in sub committee | 2 men in committee ] = 2C1*5C2 / 7C3 = 0.5714

P[ 2 women in sub committee | 3 men in committee ] = 3C1*4C2 / 7C3 = 0.5143

P[ 2 women in sub committee | 4 men in committee ] = 4C1*3C2 / 7C3 = 0.3428

P[ 2 women in sub committee | 5 men in committee ] = 5C1*2C2 / 7C3 = 0.1428

P[ 2 women in sub committee | 6 men in committee ] = 0 ( 1 women in committee )

P[ 2 women in sub committee | 7 men in committee ] = 0

P[ 2 women in sub committee | at least one men in committee ] = 0.4286*0.0542 + 0.5714*0.2089 +0.5143*0.3483 + 0.3428*0.2709 + 0.1428*0.0975 + 0.0*0.0148 + 0*0.0007 = 0.0232 + 0.1194 + 0.1791 + 0.0929 + 0.0139 + 0 + 0 = 0.4285

P[ 3 women in sub committee ] = P[ 3 women in sub committee | 1 men in committee ]*P[ 1 men in committee ] + P[ 3 women in sub committee | 2 men in committee ]*P[ 2 men in committee ] +P[ 3 women in sub committee | 3 men in committee ]*P[ 3 men in committee ] +P[ 3 women in sub committee | 4 men in committee ]*P[ 4 men in committee ] +P[ 3 women in sub committee | 5 men in committee ]*P[ 5 men in committee ] +P[ 3 women in sub committee | 6 men in committee ]*P[ 6 men in committee ] +P[ 3 women in sub committee | 7 men in committee ]*P[ 7 men in committee ]

P[ 3 women in sub committee | 1 men in committee ] = 6C3 / 7C3 = 0.5714

P[ 3 women in sub committee | 2 men in committee ] = 2C0*5C3 / 7C3 = 0.2857

P[ 3 women in sub committee | 3 men in committee ] = 3C0*4C3 / 7C3 = 0.1143

P[ 3 women in sub committee | 4 men in committee ] = 4C0*3C3 / 7C3 = 0.0286

P[ 3 women in sub committee | 5 men in committee ] = 0

P[ 3 women in sub committee | 6 men in committee ] = 0

P[ 3 women in sub committee | 7 men in committee ] = 0

P[ 3 women in sub committee | at least one men in committee ] = 0.5714*0.0542 + 0.2857*0.2089 + 0.1143*0.3483 + 0.0286*0.2709 + 0*0.0975 + 0*0.0148 + 0*0.0007 = 0.0309 + 0.0597 + 0.0398 + 0.0077 + 0 = 0.1381

P[ 0 women in sub committee | at least one men in committee ] = 0.0777

P[ 1 women in sub committee | at least one men in committee ] = 0.3496

P[ 2 women in sub committee | at least one men in committee ] = 0.4285

P[ 3 women in sub committee | at least one men in committee ] = 0.1381

Expected number of women = 0*0.0777 + 1*0.3496 + 2*0.4285 + 3*0.1381 = 0 + 0.3496 + 0.857 + 0.4143 = 1.62

This is detailed explanation of the question.

the solution used is the distribution of hyper geometric distribution, where direct results are used.