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Question: Odds are \(8\) to \(5\) against a person who is \(40\)yr old living till he is \(70\) and \(4\) to \...

Odds are 88 to 55 against a person who is 4040yr old living till he is 7070 and 44 to 33 against another person now 5050 till he will be living 8080. Probability that one of them will be alive next 3030yr. is-
(A) 5991\dfrac{{59}}{{91}}
(B) 4491\dfrac{{44}}{{91}}
(C) 5191\dfrac{{51}}{{91}}
(D) 3291\dfrac{{32}}{{91}}

Explanation

Solution

First we have to find the probability for the 2 persons to be alive and dead for the next 30 years. Later we must find the probability that one of them will be alive for the next 30 years.
In any random test or experiment, the sum of probabilities of happening and not happening an event is always equal to 11, i.e., P(E)+P(E)=1P\left( E \right) + P\left( {\overline E } \right) = 1.

Complete step-by-step answer:
Let AA be the event that person A will live in the next thirty years and let BB be the event that person B will live in the next thirty years.
We have given, odds are 88 to 55 against person A that he will live next thirty years.
So, the probability that the person A will be alive in next 3030 years, P(A)P\left( A \right) =55+8 = \dfrac{5}{{5 + 8}} =513 = \dfrac{5}{{13}}
The probability that the person A will be dead in next 3030 years, P(A)=1P(A)=1513=813P\left( {\overline A } \right) = 1 - P\left( A \right) = 1 - \dfrac{5}{{13}} = \dfrac{8}{{13}}
Now for second person B, odds are 88 to 55 against that he will live next thirty years.
So, the probability that the person B will be alive in next 3030 years,   P(B)=33+4=37\;P\left( B \right) = \dfrac{3}{{3 + 4}} = \dfrac{3}{7}
The probability that the person A will be dead in next 3030 years, P(B)=1P(B)=137=47P\left( {\overline B } \right) = 1 - P\left( B \right) = 1 - \dfrac{3}{7} = \dfrac{4}{7}
There are two ways in which one person is alive after 3030 years, i.e., ABA\overline B and AB\overline A B.
So, required probability=P(A)P(B)+P(A)P(B) = P\left( A \right) \cdot P\left( {\overline B } \right) + P\left( {\overline A } \right) \cdot P\left( B \right)
=513×47+813×37= \dfrac{5}{{13}} \times \dfrac{4}{7} + \dfrac{8}{{13}} \times \dfrac{3}{7}
=2091+2491= \dfrac{{20}}{{91}} + \dfrac{{24}}{{91}}
=4491= \dfrac{{44}}{{91}}

The required probability is =4491 = \dfrac{{44}}{{91}}.

Note: An independent event is one where the result does not get impacted by some other events. Here, ABA\overline B and AB\overline A B are two independent events, so the required probability is given by P(A)P(B)+P(A)P(B)P\left( A \right) \cdot P\left( {\overline B } \right) + P\left( {\overline A } \right) \cdot P\left( B \right).