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`
