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I am new to genetics, I have the following question at hand :

If four babies are born on a single day, what are the chances that

$A>$ Number of boys and girls will be equal.

$B>$ All $4$ will be girls.

$C>$ At least one baby will be a girl.

$D>$ Which combination of boys and girls is most likely?

My solution:

Let $X$ be the number of boys .

$X$ follows $Binomial (4,{1\over 2})$ , since the probability of a child being a boy is half.

So the answers are

$A>{3\over 8}$

$B>{1\over 16}$

$C>{15\over 16}$

$D> $ Equal number of boys and girls.

Is everything right ?

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closed as off-topic by rg255, MattDMo, WYSIWYG Oct 25 '16 at 6:55

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    $\begingroup$ I'm voting to close this question as off-topic because it's about probability, not biology.meta.biology.stackexchange.com/questions/3111/… $\endgroup$ – rg255 Oct 16 '16 at 7:15
  • $\begingroup$ @rg255 I agree. You can do so. But that doesn't mean you will downvote the question ! $\endgroup$ – Qwerty Oct 16 '16 at 13:22
  • $\begingroup$ Voting allows content that is useful and useless to the community to be identified - your question is not biology so not useful to our community, therefore I downvoted. $\endgroup$ – rg255 Oct 17 '16 at 5:13
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    $\begingroup$ @Qwerty mouse over the downvote button and read the message: "This question does not show any research effort; it is unclear or not useful." Additionally, please read "Why is voting important?". $\endgroup$ – MattDMo Oct 17 '16 at 21:21
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I know this is kind of tedious and only practical when you are considering no more than 10 children (which I think is a lot to parent), but anyway, I like to solve these problems with a binomial expansion.

So we have binomial probability so 2 options - either a boy (n) or a girl (g) at equal probability or 50% or 1/2.

(b + g)^n = a binomial expansion according to the coefficients of Pascal's triangle. Where n = 4 children, then the row of Pascal's triangle is 4 in which the first row of pascal's triangle is technically row 0.

Equation: (b + g)^4 = 1(b)^4 + 4(b^3)(g) + 6(b^2)(g^2) + 4(b)(g^3) + 1(g^4)

Problem A) So the probability that the number of boys and girls will be equal is

Probability((6(b^2)(g^2)), where (b^2)(g^2) means two boys and two girls, an equal number. When b = 1/2, and g = 1/2, or well ... b + g = 1

6(b^2)(g^2) = 6[(1/2)^2 * (1 / 2)^2] = 6 * (1/2)^4 = 6 / 16 = 3/8

B) All four will be girls: Prob((1(g^4))

= 1(1/2)^4 = 1/16

C) At least one baby will be a girl: That corresponds to 4(b^3)(g), 6(b^2)(g^2), 4(b)(g^3), and 1(g^4)

The probability that at least one will be a girl is

= (4 + 6 + 4 + 1) * (1/2)^4 = 15 / 16

D) Which combination of boys and girls is most likely? That would be 6(b^2)(g^2), so when b = g = 1/2

Prob(most likely combination) = 6/16 = 3/8 or two boys and two girls.

Note that if you add up all the combinations, the result is 100% of all probabilities:

(b + g)^4 = 1(b)^4 + 4(b^3)(g) + 6(b^2)(g^2) + 4(b)(g^3) + 1(g^4) When b = g = 1/2, then basically:

(1 / 16) + (4 / 16)+ (6 / 16) + (4 / 16) + (1 / 16) = 16 / 16 = 1 = 100%

You did it right. But I suppose this is another way to evaluate a binomial like this.

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  • $\begingroup$ A BIG typo- $2$ boys and $2$ girls, not $3$. $\endgroup$ – Qwerty Oct 16 '16 at 3:45
  • $\begingroup$ Good work, but basiacllly what you did is like - instead of directly evaluatng $3\times 4=12 $ you wrote $3+3+3+3=6+6=12$.... $\endgroup$ – Qwerty Oct 16 '16 at 3:47
  • $\begingroup$ Yeah, I was typing this so quickly! $\endgroup$ – xyz123 Oct 16 '16 at 3:48

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