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Question
The odds against student X solving a business statistics problem are 8: 6 and odds in favour of student Y solving the same problem are 14: 16 What is the chance that the problem will be solved, if they try independently?
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Solution
Let A be the event that X solves the problem B be the event that Y solves the problem.
Since the odds against student X solving the problem are 8: 6
∴ Probability of occurrence of event A is given by.
P(A) = `6/(8 + 6) = 6/14` and
P(A') = 1 – P(A) = `1 - 6/14 = 8/14`
Also, the odds in favour of student Y solving the problem are 14: 16
∴ Probability of occurrence of event B is given by
P(B) = `14/(14 + 16) = 14/30` and
P(B') = 1 – P(B) = `1 - 14/30 = 16/30`
Now A and B are independent events.
∴ A' and B' are independent events.
A' ∩ B' = Event that neither solves the problem
= P(A' ∩ B') = P(A') · P(B')
= `8/14 xx 16/30`
= `32/105`
A ∪ B = the event that the problem is solved
∴ P (problem will be solved) = P(A ∪ B)
= 1 – P(A ∪ B)`
= 1 – P(A' ∩ B')
= `1 - 32/105`
= `73/105`
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