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Question
The odds against a husband who is 60 years old, living till he is 85 are 7:5. The odds against his wife who is now 56, living till she is 81 are 5:3. Find the probability that at least one of them will be alive 25 years hence
Solution
The odds against husband living till he is 85 are 7:5.
Let P(H') = P(husband dies before he is 85)
= `7/(7 + 5) = 7/12`
So, the probability that the husband would be alive till age 85
= P(H)
= 1 – P(H')
= `1 - 7/12`
= `5/12`
Similarly, P(W') = P(Wife dies before she is 81)
Since the odds against wife will be alive till she is 81 are 5:3.
∴ P(W') = `5/(5 + 3) = 5/8`
So, the probability that the wife would be alive till age 81
= P(W)
= 1 − P(W') = `1-5/8=3/8`
Required probability = P(H ∪ W)
= P(H) + P(W) – P(H ∩ W)
Since H and W are independent events,
P(H ∩ W) = P(H).P(W)
∴ Required probability = P(H) + P(W) – P(H)P(W)
= `5/12 + 3/8 - 5/12 xx 3/8`
= `(40 + 36 - 15)/96`
= `61/96`
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