Question

The present age of a person A is 35. The odds in favour of his living upto the age 65 is 3:2. The age of another person B is 40 at present. The odds against his living upto the age of 70 is 4:1. The probability that at least one of them will be alive after 30 years is​

Answer 1

ANSWER:

13/25

GIVEN:

$\frac{p(A')}{p( A)} = \frac{3}{2} \\ \\ \frac{p(B')}{p(B)} = \frac{4}{1}$

TO FIND:

The probability that at least one of them will be alive after 30 years(p(A U B)).

FORMULAE:

$p(A \: \cup \: B) = p(A) + p(B) - p(A\: \cap \: B) \\ \\ p(A) = \frac{p(A)}{p(A') + p(A)}$

EXPLANATION:

$p(A) = \frac{2}{3 + 2} \\ \\ p(A) = \frac{2}{5} \\ \\ p(B) = \frac{1}{4 + 1} \\ \\ p(B) = \frac{1}{5} \\ \\ p(A \: \cap \: B) = p(A) \times p(B) \: because \\ A \: and \: B \: are \: independent \: events. \\ \\ p(A \: \cup \: B) = \frac{2}{5} + \frac{1}{5} - (\frac{2}{5} \times \frac{1}{5} ) \\ \\ p(A\: \cup \: B) = \frac{3}{5} - \frac{2}{25} \\ \\ If \: we \: take \: L.C.M \: we \: will \: get \\ \\ p(A \: \cup \: B) = \frac{15 - 2}{25} \\ \\ p(A \: \cup \: B) = \frac{13}{25}$

THE PROBABILITY THAT AT LEAST ONE OF THEM WILL ALIVE AFTER 30 YEARS IS 13/25.