In a family having three children, there may be no girl, one girl, two girls, or three girls. So, the probability of each is $\frac{1}{4}$ . Is this correct? Justify your answer.

Open in App

Solution

Step 1: Find the sample space
We know that Probability of an outcome $= \frac{Number of favourable outcome}{Total number of outcomes}$
There are two genders, Boy $(B)$ and Girl $(G)$ .
The total number of cases, if these $2$ genders have been taken in groups of $3$ together, $n (S) = 2^{3} = 8$ .
The cases are $B B B, B B G, B G B, B G G, G B B, G B G, G G B, G G G$ .
Step 2: Determine whether the given statement is correct or not
The cases for no girl is $B B B$ .
Thus, the number of cases for no girl is $n (G_{0}) = 1$ .
Hence, the probability of no girl is $P (G_{0}) = \frac{n (G_{0})}{n (S)} = \frac{1}{8}$ .
The cases for one girl are $B B G, B G B, G B B$ .
Thus, the number of cases for one girl is $n (G_{1}) = 3$ .
Hence, the probability of one girl is $P (G_{1}) = \frac{n (G_{1})}{n (S)} = \frac{3}{8}$ .
The cases for two girls are $B G G, G B G, G G B$ .
Thus, the number of cases for two girls is $n (G_{2}) = 3$ .
Hence, the probability of two girls is $P (G_{2}) = \frac{n (G_{2})}{n (S)} = \frac{3}{8}$ .
The cases for three girls are $G G G$ .
Thus, the number of cases for three girls is $n (G_{3}) = 1$ .
Hence, the probability of three girls is $P (G_{3}) = \frac{n (G_{3})}{n (S)} = \frac{1}{8}$ .
Hence, the probability of each case is not $\frac{1}{4}$ .
Therefore, the given statement is not correct.

Suggest Corrections

Similar questions

(a) \frac{1}{2} (b) \frac{1}{3} (c) \frac{2}{3} (d) \frac{4}{7}

\frac{1}{4}

\frac{1}{4}

4

Join BYJU'S Learning Program