# The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that at least one - Mathematics and Statistics

Sum

The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that at least one of them will be alive 20 years hence.

#### Solution

Let A be the event that husband would be alive after 20 years.
Odds against A are 8: 5
∴ The probability of occurrence of event A is given by

P(A) = 5/(8 + 5) = 5/13

∴ P(A') = 1 – P(A) = 1 - 5/13 = 8/13
Let B be the event that wife would be alive after 20 years.
Odds against B are 4: 3
∴ The probability of occurrence of event B is given by

P(B) = 3/(4 + 3) = 3/7

∴ P(B') = 1 – P(B) = 1 - 3/7 = 4/7
Since A and B are independent events
∴ A' and B' are also independent events
Let Y be the event that at least one will be alive after 20 years.
∴ P(Y) = P(at least one would be alive)
= 1 – P(both would not be alive)
= 1 – P(A' ∩ B')
= 1 – P(A') . P(B')

= 1 - 8/13 xx 4/7

= 1 - 32/91

= 59/91

Is there an error in this question or solution?
Chapter 7: Probability - Miscellaneous Exercise 7 [Page 110]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 7 Probability
Miscellaneous Exercise 7 | Q 9. (b) | Page 110
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