Question

Consider the experiment "tossing a coin twice" and probability of outcomes are as $P(H H)=\frac{1}{4}$, $P(H T)=\frac{1}{7}, P(T H)=2 / 7$ and $P(T T)=\frac{9}{28}$, then
Statement I Probability of the event $E$ : 'both the tosses yield the same result' is $\frac{4}{7}$.
Statement II Probability of the event $F$ : 'exactly two heads' is $\frac{1}{4}$.

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (B)

Let us consider an experiment of 'tossing a coin "twice". The sample space of this experiment is $S=\{H H, H T, T H, T T\}$
Let, the following probability be assigned to the outcomes
$P(H H)=\frac{1}{4}, P(H T)=\frac{1}{7}, P(T H)=\frac{2}{7}, P(T T)=\frac{9}{28}$
Clearly, this assignment satisfies the conditions of axiomatic approach. Now, let us find the probability of the event $E$ : 'Both the tosses yield the same result'.
$\text { Here, } E =\{H H, T T\}$
$\text { Now, } P(F) =\sum P\left( m _i\right), \text { for all } nn _i \in F $
$ =P(H H)+P(T T)=\frac{1}{4}+\frac{9}{28}=\frac{4}{7}$
For the event $F$ : 'exactly two heads', we have $F=\{H H\}$ and $ P(F)=P(H H)=\frac{1}{4}$