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`