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Question: In \( 20 \) years the probability that a man will be alive is \( \dfrac{3}{5} \) and the probability...

In 2020 years the probability that a man will be alive is 35\dfrac{3}{5} and the probability that his wife will be alive is 23\dfrac{2}{3} . Find the probability that at least one will be alive in 2020 years.

Explanation

Solution

Here in this question, we deal with probability. In order to solve probability related questions, we first need to understand what is probability and some of its related terms such as random experiment, outcome, sample space and many more. To solve this question, we should also have the knowledge of different properties of probability.

Complete step by step solution:
Probability helps us to know how often an event is likely to happen.
Some of the properties of probability are:
P(A)+P(A)=1P\left( A \right) + P\left( {A'} \right) = 1
P(AB)=P(A)+P(B)P(AB)P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)
P(AB)=P(A)+P(B)P(AB)P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)
P(A1A2A3...An)=P(A1)+P(A2)+....+P(An)P\left( {{A_1} \cup {A_2} \cup {A_3}... \cup {A_n}} \right) = P\left( {{A_1}} \right) + P\left( {{A_2}} \right) + .... + P\left( {{A_n}} \right)
Now, according to the question
Let A be the event where the husband will be alive in 2020 years and B be the event where the wife will be alive in 2020 years. As we can clearly see that A and B are independent events.
Given is P(A)=35,P(B)=23P\left( A \right) = \dfrac{3}{5},P\left( B \right) = \dfrac{2}{3}
The probability that at least one of them will be alive in 2020 years is given by
P(AB)=P(A)+P(B)P(AB)P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) ----(i)
The value of P(AB)P\left( {A \cap B} \right) can be calculated by P(AB)=P(A)P(B)P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right) . Using the same as well as given values in equation (i) we get,
P(AB)=P(A)+P(B)P(AB) P(A)+P(B)P(A)(B) 35+23(35)(23) 35+2325 15+23 3+1015=1315  \Rightarrow P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \\\ \Rightarrow P\left( A \right) + P\left( B \right) - P\left( A \right)\left( B \right) \\\ \Rightarrow \dfrac{3}{5} + \dfrac{2}{3} - \left( {\dfrac{3}{5}} \right)\left( {\dfrac{2}{3}} \right) \\\ \Rightarrow \dfrac{3}{5} + \dfrac{2}{3} - \dfrac{2}{5} \\\ \Rightarrow \dfrac{1}{5} + \dfrac{2}{3} \\\ \Rightarrow \dfrac{{3 + 10}}{{15}} = \dfrac{{13}}{{15}} \\\

Hence, the required probability is 1315\dfrac{{13}}{{15}} .

Note: A random experiment is a trial or an experiment whose outcomes cannot be predicted with surety or certainty. Outcome is the result of any random experiment conducted. Sample space is a set of all the possible outcomes for a random experiment. Event is a set of possible outcomes under a specified condition.