Question

Find the probability to 3 decimal places that when a couple has three children, at least one of them is a boy. (Assume that boys and girls are equally likely.)

Answer 1

Answer:

0.875

Step-by-step explanation:

This is a binomial distribution because a child can be either a boy or a girl. We denote the probability of being a boy as p and being a girl as q. Both are mutually exclusive. The questions both are equally likely. Hence, p = q = 0.5.

The event of having at least a boy is the complement of the event of having no boy. Now the probability of having no boy is the the probability of all children being girls. This is given by

$P(3G) = 0.5\times0.5\times0.5 = 0.125$

Then, the probability of at least 1 boy, by the law of complements, is

$P(\ge1B) = 1 - P(3G) = 1 - 0.125 = 0.875$

Answer 2

Answer:

The probability of having at least one is a boy = 0.875

Step-by-step explanation:

Let B represent boy and G represent Girl.

For a couple having three children with the probability of having a boy or a girl is the same, they are 8 possible outcomes which are [BBB, BBG, BGG, BGB, GGG, GGB, GBG, GBB] = 8

BBB means having a boy followed by a boy followed by a boy while BGB means having a boy followed by a girl followed by a boy.

The probability of having at least one is a boy, it can be [ BBB BBG BGG BGB GGB GBG GBB] = 7  

Probability is the ratio number of favorable outcomes to the total number of possible outcomes.

Therefore, The probability of having at least one is a boy = 7/8 = 0.875