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 $P(X=k)=\binom{n}{k}p^k (1-p)^{n-k}$
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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