Digital-Logic-Design
October 5, 2023GATE 2004-IT
October 5, 2023GATE 2004-IT
| Question 1 |
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?
| 3/23 | |
| 6/23 | |
| 3/10 | |
| 3/5 |
Question 1 Explanation:
Let us consider total no. of families = 100
In that 50% of families having 3 children
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 children = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23
In that 50% of families having 3 children
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 children = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23
Correct Answer: B
Question 1 Explanation:
Let us consider total no. of families = 100
In that 50% of families having 3 children
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 children = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23
In that 50% of families having 3 children
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 children = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23