The odds against a person who is 40 years old | vedclass

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(B) Let $A$ be the event that the first person is alive after $30$ years and $B$ be the event that the second person is alive after $30$ years.Given t

Vedclass · Accepted Answer

$\frac{44}{91}$

Vedclass · Answer

(B) Let $A$ be the event that the first person is alive after $30$ years and $B$ be the event that the second person is alive after $30$ years.Given the odds against $A$ are $8:5$,the probability that the first person dies is $P(\bar{A}) = \frac{8}{8+5} = \frac{8}{13}$.Thus,the probability that the first person is alive is $P(A) = 1 - \frac{8}{13} = \frac{5}{13}$.Given the odds against $B$ are $4:3$,the probability that the second person dies is $P(\bar{B}) = \frac{4}{4+3} = \frac{4}{7}$.Thus,the probability that the second person is alive is $P(B) = 1 - \frac{4}{7} = \frac{3}{7}$.The event that exactly one of them is alive is $(\bar{A} \cap B) \cup (A \cap \bar{B})$.Since these events are independent,the probability is $P(\bar{A})P(B) + P(A)P(\bar{B})$.$= (\frac{8}{13} 	imes \frac{3}{7}) + (\frac{5}{13} 	imes \frac{4}{7})$$= \frac{24}{91} + \frac{20}{91} = \frac{44}{91}$.

The odds against a person who is $40$ years old living until he is $70$ are $8$ to $5$,and the odds against another person who is $50$ years old living until he is $80$ are $4$ to $3$. Find the probability that exactly one of them will be alive $30$ years from now.

Similar Questions

Fill in the blank in the following table:

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$0.35$ $..........$ $0.25$ $0.6$

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$A$ and $B$ are two events such that $P(A)=0.54$,$P(B)=0.69$ and $P(A \cap B)=0.35$. Find $P(A \cup B)$.

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