In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?
(A) 3/23
(B) 6/23
(C) 3/10
(D) 3/5
Answer: (B)
Explanation: x- total families. Given:
- 50% -3 Children -> (5*x/10) * 3=(1*x/2)*3
- 30% -2 Children-> (3*x /10) * 2
- Rear 20%- 1child -> (2*x/10)*1 =1*x/5
Probability = No of ways it can happen/Total Outcomes
=( (3/10)*2 ) / ( (1/2)*3 + (3/10)*2 +1/5 )
= 6/23
So Answer is B
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