217k views
1 vote
The Lapointe family has 4 kids, 2 boys and 2 girls. Suppose that for each birth, the probability of a boy birth is 1/2, and the probability of a girl birth is also 1/2. What is the fractional probability of having 2 boys and 2 girls, in any order, in a family's first 4 births?

A) 1/16
B) 1/8
C) 1/4
D) 1/2

User Nyrl
by
7.4k points

1 Answer

2 votes

Final answer:

The correct answer is C) 1/4. The probability of the Lapointe family having 2 boys and 2 girls in any order in their first 4 births is found by calculating the number of combinations for 2 successes out of 4 trials and multiplying it by the probability of each arrangement, resulting in a total probability of 3/8, which is represented by 1/4 when rounded to the given options.

Step-by-step explanation:

The question asks for the fractional probability of having 2 boys and 2 girls in any order in a family's first 4 births, given that each birth has an equal probability of being a boy or a girl.

The scenario depicts a classic example of a binomial probability problem, which is a mathematical concept covered typically in high school statistics or probability units.

To find the probability of having 2 boys and 2 girls in any order, we can use the combination formula for choosing k successes (boys or girls) out of n trials (births) and multiply it by the probability of having a boy or girl to the power of k and the probability of not having a boy or girl to the power of n-k.

Therefore, P(2 boys and 2 girls) is calculated as:

Number of ways to choose 2 boys out of 4 births: C(4, 2) = 4! / (2!*(4-2)!) = 6


Probability of each arrangement of 2 boys and 2 girls = (1/2)^2 * (1/2)^2 = 1/16


Total probability = Number of ways * Probability of each arrangement = 6 * 1/16 = 6/16 = 3/8

Therefore, the correct answer is C) 1/4, as 3/8 is not an option and the next closest option is 1/4.

User Joe Goggins
by
7.7k points