Question

Mr A has 6 children and atleast one child is a girl ,then probability that Mr A has 3 boys and 3 girls is?

Answer 1

$\frac{1}{6}$

Answer 2

Answer:

$\dfrac{20}{63}$

Step-by-step explanation:

Please note that this question is based on conditional probability, the condition being, "at least one child is a girl".

Let A and B are events defined as follows:

$A \equiv \text{There are 3 boys and 3 girls}\\B \equiv \text{Atleast one child is a girl}$

From conditional probability,

$P(A | B) = \dfrac{P(A \displaystyle \cap B)}{P(B)}$

In this case, $P(A \displaystyle \cap B) = P(\text{(There are 3 boys and 3 girls) and (there is at least one girl)}) = P(A)$

Therefore,

$P(A | B) = \dfrac{P(A)}{P(B)}$

$P(A) = \text{Probability of 3 boys + 3 girls}$

Out of the six childern, choose any three as boys. The remaining are girls.

There are $\displaystyle{6 \choose 3}$ ways for this.

Total no. of possible combinations of boys and girls = $2^6$

$P(A) = \dfrac{\displaystyle{6 \choose 3}}{2^6} = \dfrac{20}{64}$

$P(B) = P(\text{at least one girl}) = 1 - P(\text{no girl}) = 1 - \dfrac{1}{2^6} = \dfrac{63}{64}$

Therefore, required probability = $P(A | B) = \dfrac{P(A)}{P(B)} = \dfrac{\frac{20}{64}}{\frac{63}{64}} = \dfrac{20}{63}$

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