Question

Assuming boys and girls are equally​ likely, find the probability of a couple having a baby boyboy when their sixthsixth child is​ born, given that the first fivefive children were all boysall boys.

Answer 1

Doesn't matter what the first five are. The probability is always 1/2 for the next child. This is like tossing a coin. If you toss a coin 5 times and all the times are tails, what is the probability that the sixth time will be heads. If it's a fair coin, it's 1/2.

Answer 2

Answer:

50% probability of a couple having a baby boyboy when their sixth child is​ born, given that the first five were all boys.

Step-by-step explanation:

For each child, there are only two possible outcomes. Either it is a boy, or it is a girl. The probability of a child being a boy or a girl is independent of the other children. So we use the binomial probability distribution to solve this question.

The conditional probability formula is also used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Find the probability of a couple having a baby boyboy when their sixth child is​ born, given that the first five children were all boys.

Event A: First five children being boys.

Event B: Sixth children being a boy.

P(A):

First five children being boys.

P(X = 5) when n = 5.

Equally likely to be boy or girl, so p = 0.5.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

Intersection:

Intersection between the first five children being boys and the sixth also being a boy, so $P(X = 6)$ when n = 6.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A \cap B) = P(X = 6) = C_{6,6}.(0.5)^{6}.(0.5)^{0} = 0.015625$

Probability:

$P(B|A) = \frac{0.015625}{0.03125} = 0.5$

50% probability of a couple having a baby boyboy when their sixth child is​ born, given that the first five were all boys.