Question

Assertion :

Consider a bag containing $3$ normal and $6$ doubly headed coins. A coin is selected randomly and tossed $3$ times and found to fall head-wise on all the $3$ occasions.

Probability that the selected coin was a normal coin is $\frac{1}{17}$. Reason: Probability that the selected coin was a doubly headed coin is $\frac{15}{16}$.

• 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. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Probability that the coin was a normal coin is

$=\frac{\frac{3}{9}\frac{1}{2^3}}{\frac{3}{9}\frac{1}{2^3}+\frac{6}{9}\left(1\right)}$

$=\frac{\frac{1}{24}}{\frac{1}{24}+\frac{2}{3}}$

$=\frac{1}{1+\frac{24\times 2}{3}}$

$=\frac{1}{17}$.

Probability that it was a double headed coin is

$=\frac{\frac{6}{9}\left(1\right)}{\frac{3}{9}\frac{1}{2^3}+\frac{6}{9}\left(1\right)}$

$=\frac{\frac{2}{3}}{\frac{1}{24}+\frac{2}{3}}$

$=\frac{16}{1+16}$

$=\frac{16}{17}$.

Hence assertion is correct while reason is incorrect.