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

Question

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

Tardigrade Admin · Accepted Answer

Let us consider an experiment of 'tossing a coin "twice". The sample space of this experiment is

S = {HH, H T, T H, TT}

Let, the following probability be assigned to the outcomes

P (HH) = \frac{1}{4}, P (H T) = \frac{1}{7}, P (T H) = \frac{2}{7}, P (TT) = \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'.

Here, E = {HH, TT}

Now, P (F) = \sum P (m_{i}), for all n n_{i} \in F

= P (HH) + P (TT) = \frac{1}{4} + \frac{9}{28} = \frac{4}{7}

For the event

F

: 'exactly two heads', we have

F = {HH}

and

P (F) = P (HH) = \frac{1}{4}