Question

A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. None of these.

Answer 1

The correct option is A

$\frac{3}{4}$

Explanation for correct options:

Option (A): $\frac{3}{4}$

We know,

Probability = $\frac{Favourableevents}{Totalnumberofevents}$

Probability of getting a head (p) = $\frac{1}{2}$

The Probability of a five or six (q) = $\frac{2}{6}$

= $\frac{1}{3}$

The required probability = $p+(1-p)(1-q)p+(1-p)(1-q)(1-p)(1-q)....$

= $\frac{1}{2}+(1-\frac{1}{2})(1-\frac{1}{3})\frac{1}{2}+(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{2})(1-\frac{1}{3})\frac{1}{2}+...$

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

= $\frac{1}{2}\left(\frac{1}{1-{\displaystyle \frac{1}{3}}}\right)$

= $\frac{3}{4}$

Therefore, the probability is $\frac{3}{4}$.

Explanation for incorrect options:

Option (B): $\frac{1}{2}$

Here, the probability of getting a head in the coin before getting a five or six on the dice = $\frac{3}{4}$

Therefore, the probability is not $\frac{1}{2}$.

Option (C): $\frac{1}{3}$

Here, the probability of getting a head in the coin before getting a five or six on the dice = $\frac{3}{4}$

Therefore, the probability is not $\frac{1}{3}$

Option (D): None of these

Here, the probability of getting a head in the coin before getting a five or six on the dice = $\frac{3}{4}$

Therefore, probability is one of these options.

Hence, Option (A) is the correct answer.