Question

If ${{\text{E}}_1}$ and ${{\text{E}}_2}$ are two events such that P (${{\text{E}}_1}$) = 1/4, P (${{\text{E}}_2}$ / ${{\text{E}}_1}$) = 1/2 and P (${{\text{E}}_1}$ / ${{\text{E}}_2}$) = 1/4, then
A. ${{\text{E}}_1}$ and ${{\text{E}}_2}$ are independent.
B. ${{\text{E}}_1}$ and ${{\text{E}}_2}$ are exhaustive.
C. ${{\text{E}}_2}$ is twice as likely to occur as ${{\text{E}}_1}$.
D. probabilities of the events ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$, ${{\text{E}}_1}$ and ${{\text{E}}_2}$ are in G.P.

Answer 1

Hint: To solve this question we will use the property of probabilities and then check all the given options to find which one is correct.

Complete step-by-step answer:
Now, if there are two events A and B then P (A / B) = P ( ${\text{A }} \cap {\text{ B}}$) / P (B). We will use this property to find the correct option(s).
Now ,using the given conditions in the question, P ( ${{\text{E}}_1}$ ) = 1/4 , P ( ${{\text{E}}_2}$ / ${{\text{E}}_1}$ ) = 1/2
P ( ${{\text{E}}_2}$ / ${{\text{E}}_1}$ ) = P ( ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$) /P ( ${{\text{E}}_1}$) = 1/2
$\dfrac{{{\text{P ( }}{{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}{\text{ )}}}}{{\dfrac{1}{4}}}{\text{ = }}\dfrac{1}{2}$
Therefore,
Also, P (${{\text{E}}_1}$ / ${{\text{E}}_2}$) = 1/4. So, applying the property, we get
P (${{\text{E}}_1}$ / ${{\text{E}}_2}$) = P ( ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$) /P (${{\text{E}}_2}$) = 1/4
$\dfrac{{\dfrac{1}{8}}}{{{\text{P ( }}{{\text{E}}_2})}}{\text{ = }}\dfrac{1}{4}$
${\text{P ( }}{{\text{E}}_2}){\text{ = }}\dfrac{1}{2}$
Now, we will check all the given options one by one to find which are correct.
Taking option A. Now, two events are independent if the product of their probability is equal to the probability of their intersection. So,
P ( ${{\text{E}}_1}$). P ( ${{\text{E}}_2}$) = $\dfrac{1}{4}{\text{ }} \times {\text{ }}\dfrac{1}{2}{\text{ = }}\dfrac{1}{8}$
Also, P ( ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$) = $\dfrac{1}{8}$
So, both the events are independent. So, option (A) is correct.
Now, checking option B. The events are called exhaustive if the sum of their probability is equal to 1.
So, P ( ${{\text{E}}_1}$) + P ( ${{\text{E}}_2}$) = $\dfrac{1}{4}{\text{ + }}\dfrac{1}{2}{\text{ = }}\dfrac{3}{4}$
So, the sum of given probabilities is not equal to 1. So, option (B) is incorrect.
No, checking option C.
2 times the probability of event ${{\text{E}}_1}$ = 2 x P ( ${{\text{E}}_1}$) = $2{\text{ }} \times \dfrac{1}{4}{\text{ = }}\dfrac{1}{2}$ = P ( ${{\text{E}}_2}$).
So, option (C) is correct.
Now, checking the last option i.e. option (D).
P ( ${{\text{E}}_1}$) = $\dfrac{1}{4}$, P ( ${{\text{E}}_2}$) = $\dfrac{1}{2}$, P ( ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$) = $\dfrac{1}{8}$. So, probabilities of ${{\text{E}}_1}{\text{ }} \cap {\text{ }}{{\text{E}}_2}$, ${{\text{E}}_1}$, ${{\text{E}}_2}$ are in G.P with a common ratio of $\dfrac{1}{2}$.
So, option (D) is correct.
So, except option (B), all the other options are correct.

Note: While solving questions of probability make sure that you use the correct formula. Many students make silly mistakes by reversing the terms of numerator and denominator. Also, carefully do all the calculation while finding probability.