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Question
Students of under graduation submitted a case study on “Understanding the Probability of Left-Handedness in Children Based on Parental Handedness”. Following Recent studies suggest that roughly 12% of the world population is left-handed. Depending on the parents’ handedness, the chances of having a left-handed child are as follows:
Scenario A: Both parents are left-handed, with a 24% chance of the child being left-handed.
Scenario B: The fathers is right-handed and the mothers left-handed, with a 22% chance of child being left-handed.
Scenario C: The fathers left-handed and the mother is right-handed, with a 17% chance of child being left-handed.
Scenario D: Both parents are right-handed, with a 9% chance of having a left-handed child.
Assuming that scenarios A, B, C and D are equally likely and L denotes the event that the child is left-handed, answer the following questions.
- What is the overall probability that a randomly selected child is left-handed?
- Given that exactly one parent is left-handed, what is the probability that a randomly selected child is left-handed?
- If a child is left-handed, what is the probability that both parents are left-handed?
Solution
a. Since events A, B, C, D are equally likely
`\implies P(A) = P(B) = P(C) = P(D) = 1/4`
As per question,
P(L/A) = `24/100`,
P(L/B) = `22/100`,
P(L/C) = `17/100`,
P(L/D) = `9/100`
The probability that a randomly selected child is left-handed
= P(A) × P(L/A) + P(B) × P(L/B) + P(C) × P(L/C) + P(D) × P(L/D) P(A/L)
= `1/4 xx 24/100 + 1/4 xx 22/100 + 1/4 xx 17/100 + 1/4 xx 9/100`
b. The probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed
= P(L/B) + P(L/C)
= `22/100 + 17/100`
= `39/100`
c. P(A/L)
= `(P(A) xx P(L//A))/(P(A) xx P(L//A) + P(B) xx P(L//B) + P(C) xx P(L//C) + P(D) xx P(L//D)P(A//L)`
= `(1/4 xx 24/100)/(1/4 xx 24/100 + 1/4 xx 22/100 + 1/4 xx 17/100 + 1/4 xx 9/100)`
= `1/3`
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