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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?

  1. 3/23
  2. 6/23
  3. 3/10
  4. 3/5
asked in Probability by Veteran (19.4k points)   | 733 views

2 Answers

+8 votes
Best answer
Answer is B) 6/23

Let N be the total number of families.

Number of children in a  family of 3 children = (N/2) * 3

Number of children in a  family of 2 children = (3N /10) * 2

Number of children in a  family of 1 child = (N/5) * 1

Probability = Favorable case / Total cases

               =( (3/10)*2 ) / ( (1/2)*3 + (3/10)*2 +1/5 )

               = 6/23
answered by Active (2.1k points)  
selected by
@ Prateeksha ma'am ,
Why Bayes' theorem is not applied here???
Exactly..

Plz someone solve using Bayes' theorem..
What is the requirement for conditional probability in this question?
+1 vote

Suppose,

50 families have three children(150 child's),

30 families have two children(60 childs)

and remaining 20 families have one child(20 childs).

So ,Probability that a randomly picked child belongs to a family with two children =

$\frac{60}{150+60+20}$=$\frac{60}{230}$=$\frac{6}{23}$

answered by Veteran (17k points)  


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