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A family has n children with probability αpn,n≥1whereα≤(1-p)/p

(a) What proportion of families has no children?

(b) If each child is equally likely to be a boy or a girl (independently of each other), what proportion of families consists of k boys (and any number of girls)?

Short Answer

Expert verified

The answer of part (a) is

PN=0=1-αp1-p

Part (b) is Pkboys=∑n=k∞nk12nαpn

Step by step solution

01

Step 1:Given Information(Part-a)

The random variable Nthat denotes the number of children . we know that N∈N0and thatP(N=n)=αpn.for everyn≥1.we are required to find outPN=0.

02

Step 2:Calculation (Part-a)

PN=0=1-P(N≥1)

= 1-∑k=1αpn

= 1-αp1-p

03

Step 3:Final Answer(Part-a)

The answer isPN=0=1-αp1-p

04

Step 4:Given Information(Part-b)

If we are given that some family has nchildren, the number of boys in that family has binomial distribution with parameters nand1/2.

05

Step 5:Calculation (Part-b)

Pkboys=∑n=k∞Pkboys/N=nPN=n

=∑n=k∞nk12nαpn

06

Step 6:Final Answer (Part-b)

The answer is P(k boys) =∑n=k∞nk12nαpn

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