Three students A, B and C can solve 50%, 60% and 70% sums from a book. If one sum from that book is given then to solve, then what is the probability that the problem will be solved?
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P(A) = (50/100) = (5/10),
P(B) = (60/100) = (6/10) and
P(C) = (70/100) = (7/10)
A, B, C are independent events. ∴ P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)'
= 1 – P(A') ∙ P(B') ∙ P(C') =
1 – [(5/10)(4/10)(3/10)] =
1 – (6/100) =
0.94 ans
Answered by
1
Answer:
The correct answer 0.94
Explanation:
P(A) = (50/100) = (5/10), P(B) = (60/100) = (6/10) and P(C) = (70/100) = (7/10)
A, B, C are independent events.
∴ P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)'
= 1 – P(A') ∙ P(B') ∙ P(C')
= 1 – [(5/10)(4/10)(3/10)]
= 1 – (6/100) = 0.94
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