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प्रश्न
A die is thrown twice. Let A be the event, ‘First die shows 5’ and B be the event, ‘second
die shows 5’. Find P(A ∪ B)
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उत्तर
A die is thrown twice.
Let S be the sample space
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3 ,2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event ‘First die shows 5’
A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,0)}
n(A) = 6
P(A) = `("n"("A"))/("n"("S"))`
= `6/36`
= `1/6`
Let B be the event ‘Second die shows 5’
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6
P(B) = `("n"("B"))/("n"("S"))`
= `6/36`
= `1/6`
A ∩ B = {(5, 5)}
n(A ∩ B) = 1
P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S"))`
= `1/36`
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= `1/6 + 1/6 - 1/36`
= `(6 + 6 - 1)/36`
= `11/36`
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