Question

The probability that Krishna will be alive 10 years hence, is $${\displaystyle \frac{7}{15}}$$ and that Hari will be alive is $${\displaystyle \frac{7}{10}}$$. The probability both Krishna and Hari will be dead 10 years. Hence is:
A.$${\displaystyle \frac{21}{150}}$$
B.$${\displaystyle \frac{24}{150}}$$
C.$${\displaystyle \frac{49}{150}}$$
D.$${\displaystyle \frac{56}{150}}$$

Answer 1

Hint: Here, we will first assume that A be the event when Krishna will be alive and B be the event when Hari will be alive. Then we will use the property of probability, $$P\left(A^c\right)=1-P\left(A\right)$$, where $$P\left(A^c\right)$$ is complement of $$P\left(A\right)$$that is, Krishna will be dead and property of probability, $$P\left(B^c\right)=1-P\left(B\right)$$, where $$P\left(B^c\right)$$ is complement of $$P\left(B\right)$$, that is, Hari will be dead. Then since both the events are independent of each other, we have $$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$$ and getting $$P\left(A^c\cap B^c\right)=P\left(A^c\right)P\left(B^c\right)$$ to find the required value.

Complete step-by-step answer:
Let us assume that A be the event when Krishna will be alive and B be the event when Hari will be alive.
So we given that the probabilities that Krishna will be alive 10 years hence, $$P\left(A\right)={\displaystyle \frac{7}{15}}$$ and that Hari will be alive is $$P\left(B\right)={\displaystyle \frac{7}{10}}$$.
Since we know that the property of probability, $$P\left(A^c\right)=1-P\left(A\right)$$, where $$P\left(A^c\right)$$ is complement of $$P\left(A\right)$$that is, Krishna will be dead, we get
$$\begin{gathered} \Rightarrow P\left(A^c\right)=1-{\displaystyle \frac{7}{15}} \\ \Rightarrow P\left(A^c\right)={\displaystyle \frac{15-7}{15}} \\ \Rightarrow P\left(A^c\right)={\displaystyle \frac{8}{15}} \end{gathered}$$
Using the property of probability, $$P\left(B^c\right)=1-P\left(B\right)$$, where $$P\left(B^c\right)$$ is complement of $$P\left(B\right)$$, that is, Hari will be dead, we get

$$\begin{gathered} \Rightarrow P\left(B^c\right)=1-{\displaystyle \frac{7}{10}} \\ \Rightarrow P\left(B^c\right)={\displaystyle \frac{10-7}{10}} \\ \Rightarrow P\left(B^c\right)={\displaystyle \frac{3}{10}} \end{gathered}$$

Since both the events are independent of each other, we have $$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$$.
So now substituting the values of $$P\left(A^c\right)$$and $$P\left(B^c\right)$$in the property of probability,$$P\left(A^c\cap B^c\right)=P\left(A^c\right)P\left(B^c\right)$$, we get
$$\begin{gathered} \Rightarrow P\left(A^c\cap B^c\right)={\displaystyle \frac{8}{15}}\times {\displaystyle \frac{3}{10}} \\ \Rightarrow P\left(A^c\cap B^c\right)={\displaystyle \frac{24}{150}} \end{gathered}$$
Hence, option A is correct.


Note: Students must know that they have to assume the given conditions with some distinct variables to understand the question properly. In solving these types of questions, students must know about basic properties of probability, $$P\left(A^c\right)=1-P\left(A\right)$$ and since both the events are independent of each other, we have $$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$$.