# 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 wi - Mathematics and Statistics

Sum

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

Concept: Odds (Ratio of Two Complementary Probabilities)
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