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Question
Read the following passage:
|
Recent studies suggest the roughly 12% of the world population is left-handed.
Assuming that P(A) = P(B) = P(C) = P(D) = `1/4` and L denotes the event that child is left-handed. |
Based on the above information, answer the following questions:
- Find `P(L/C)` (1)
- Find `P(overlineL/A)` (1)
- (a) Find `P(A/L)` (2)
OR
(b) Find the probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed. (2)
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Solution
Given, P(L) = `12/100`
`P(L) P(overlineL) = 1 - 12/100 = 88/100`
and P(A) = P(B) = P(C) = P(D) = `1/4`
`P(L/A) = 24/100`,
`P(L/B) = 22/100`,
`P(L/C) = 17/100`,
`P(L/D) = 9/100`
(i) `P(L/C) = 17/100`, from the given data.
(ii) `P(overlineL/A) = (P(overlineL ∩ A))/(P(A))`
= `(P(A) - P(L ∩ A))/(P(A))`
= `1 - (P(L ∩ A))/(P(A))`
= `1 - P(L/A)`
= `1 - 24/100`
= `(100 - 24)/100`
= `76/100`
= `38/50`
= `19/25`
(iii) (a) `P(A/L) = (P(A ∩ L))/(P(L))`
But `P(L/A) = (P(A ∩ L))/(P(A))`
`24/100 = (P(A ∩ L))/(1/4)`
`\implies` P(A ∩ L) = `24/100 xx 1/4 = 6/100 = 3/50`
∴ `P(A/L) = (3/50)/(12/100) = (3 xx 100)/(12 xx 50) = 1/2`.
OR
(b) `P(L/(B ∪ C)) = (P[(L) ∩ (B ∪ C)])/(P(B ∪ C))`
= `(P[(L ∩ B) ∪ (L ∩ C)])/(P(B ∪ C))`
= `(P(L ∩ B) + P(L ∩ C) - P(L ∩ B)P(L ∩ C))/(P(B) + P(C) - P(B)P(C))` ...(As they are independent)
= `(22/100 xx 1/4 + 17/100 xx 1/4 - 22/400 xx 17/400)/(1/4 + 1/4 - 1/4 xx 1/4)`
= `(22/400 + 17/400 - (22 xx 17)/(400 xx 400))/(1/2 - 1/16)`
= `((39/400 - 374/160000))/(1/2 - 1/16)`
= `16/7((39 xx 400 - 374)/160000)`
= `(16 xx 15226)/(7 xx 160000)`
= 0.217
= 0.22
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