Solve the problem.
-Assume that the probability is $\frac { 1 } { 2 }$ that a child born is a boy. What is the probability that of the four children in a family two are girls and two are boys?
A) $\frac { 1 } { 8 }$
B) $\frac { 3 } { 4 }$
C) $\frac { 1 } { 2 }$
D) $\frac { 1 } { 4 }$

Question

Solve the problem.
-Assume that the probability is $\frac { 1 } { 2 }$ that a child born is a boy. What is the probability that of the four children in a family two are girls and two are boys?
A)  $\frac { 1 } { 8 }$
B)  $\frac { 3 } { 4 }$
C)  $\frac { 1 } { 2 }$
D)  $\frac { 1 } { 4 }$

Quizplus · Accepted Answer

The probability of having two girls and two boys in any order in a family of four children can be calculated using the binomial probability formula, which in this case is $ \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = 6 \times \frac{1}{4} \times \frac{1}{4} = \frac{6}{16} = \frac{3}{8} $. However, there seems to be a mistake in my initial calculation and explanation. The correct calculation should indeed reflect the scenario of having two boys and two girls out of four children, considering the probability of having a boy or a girl is $\frac{1}{2}$.Correcting the oversight: The number of ways to have two boys and two girls in a family of four children, given that each child is equally likely to be a boy or a girl, is represented by the binomial coefficient $ \binom{4}{2} $, which equals 6. Since the probability of each child being a boy or a girl is $\frac{1}{2}$, and since these events are independent, the probability of any specific sequence of two boys and two girls is $ \left(\frac{1}{2}\right)^4 = \frac{1}{16} $. Multiplying the number of such sequences (6) by the probability of each sequence gives $ 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} $, indicating an error in the provided options and my initial response.Given the error in the initial response and the options provided, the correct probability calculation was inaccurately matched to the available choices. The accurate probability of having two boys and two girls in a family of four, with each child having a $\frac{1}{2}$ chance of being a boy or a girl, is indeed $\frac{3}{8}$, which does not match any of the options provided (A, B, C, D). This indicates a mistake in my explanation regarding the matching of the correct probability to the given options.

Solve the Problem 12\frac { 1 } { 2 }21​ That a Child Born Is a Boy

Solve the Problem $\frac { 1 } { 2 }$ That a Child Born Is a Boy