What Is the Probability of Having Exactly k Boys in a Family of n Children?

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Discussion Overview

The discussion revolves around calculating the probability of having exactly k boys in a family of n children, given a probability p for each child being a boy. The scope includes mathematical reasoning related to probability theory.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the probability is given by the formula $P(X=k)=p^k \cdot (1-p)^{n-k}$.
  • Another participant agrees with this initial proposal, affirming its correctness.
  • A later reply challenges the initial formula, suggesting that the correct probability should include a binomial coefficient, stating $P(X=k)=\binom{n}{k}p^k (1-p)^{n-k}$.

Areas of Agreement / Disagreement

There is disagreement regarding the correct formula for the probability, with one participant supporting the initial claim and another providing a different formula that includes a binomial coefficient.

Contextual Notes

The discussion does not clarify the assumptions regarding the independence of births or the specific values of n and p, which may affect the application of the proposed formulas.

evinda
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Hello! (Wave)

A couple gets $n$ children. At each birth, the probability to get a boy is $p$ (independent births). Which is the probability that exactly $k$ of the children are boys?

I have thought the following:

Let $X$ be the number of boys that the couple gets. Then the desired probality is

$P(X=k)=p^k \cdot (1-p)^{n-k}$

Am I right? (Thinking)
 
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Hey evinda!

Yes, that is correct. (Nod)
 
I like Serena said:
Hey evinda!

Yes, that is correct. (Nod)

Great! Thank you (Happy)
 
evinda said:
Hello! (Wave)

A couple gets $n$ children. At each birth, the probability to get a boy is $p$ (independent births). Which is the probability that exactly $k$ of the children are boys?

I have thought the following:

Let $X$ be the number of boys that the couple gets. Then the desired probality is

$P(X=k)=p^k \cdot (1-p)^{n-k}$

Am I right? (Thinking)
Hello,

Your answer should be $P(X=k)=\binom{n}{k}p^k (1-p)^{n-k}$
 

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