Question

Assertion :

A fair coin is being tossed four times.

Consider the following events:

A is the event all four results are the same.

B is the event exactly one Head occurs.

C is the event at least two Heads occur.

$P\left(A\right)+P\left(B\right)+P\left(C\right)>1$ Reason: A, B and C are not Mutually Exclusive since events A and C have outcomes in common.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Given:

coin tossed four times.

A is the event all four results are the same.

B is the event exactly one Head occurs.

C is the event at least two Heads occur.

To find:

p(A)

The sample space for tossing of fair coin four times is,

{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} = 16 ways.

$p\left(A\right)=\frac{\text{number of all four results are same}}{\text{Total number of outcomes}}=\frac{2}{16}=\frac{1}{8}$

$p\left(B\right)=\frac{\text{number of exactly one head}}{\text{Total number of outcomes}}=\frac{4}{16}=\frac{1}{4}$

$p\left(C\right)=\frac{\text{number events of atleast two head occurs}}{\text{Total number of outcomes}}=\frac{11}{16}$

$p\left(A\right)+p\left(B\right)+p\left(C\right)=\frac{2}{16}+\frac{4}{16}+\frac{11}{16}=\frac{17}{16}>1$