Question

If a couple plans to have 8 ​children, what is the probability that there will be at least one boy​? Assume boys and girls are equally likely. Is that probability high enough for the couple to be very confident that they will get at least one boy in 8 ​children?The probability is ___ . ?(Type an integer or a simplified? fraction.

Answer 1

Answer:

99.61% probability that there will be at least one boy, which is high enough for the couple to be very confident that they will get at least one boy in 8 ​children.

The probability is 0.9961

Step-by-step explanation:

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

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.

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

Assume boys and girls are equally likely.

This means that $p = 0.5$

If a couple plans to have 8 ​children, what is the probability that there will be at least one boy​?

This is $P(X > 0)$ when $n = 8$

We know that either there are no boys, or there is at least one boy. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X > 0) = 1$

$P(X > 0) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{8,0}.(0.5)^{0}.(0.5)^{8} = 0.0039$

$P(X > 0) = 1 - P(X = 0) = 1 - 0.0039 = 0.9961$

Any probability above 95% is considered very high.

99.61% probability that there will be at least one boy, which is high enough for the couple to be very confident that they will get at least one boy in 8 ​children.