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

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The discussion centers on calculating the probability of having exactly k boys in a family with n children, where the probability of each child being a boy is p. The initial formula proposed was P(X=k)=p^k * (1-p)^(n-k), which is incorrect. The correct formula for this probability is P(X=k)=C(n,k) * p^k * (1-p)^(n-k), where C(n,k) is the binomial coefficient representing the number of ways to choose k boys from n children. This correction emphasizes the importance of including the combinatorial factor in the probability calculation.
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 $P(X=k)=\binom{n}{k}p^k (1-p)^{n-k}$
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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