A man alternatively tosses a fair coin and rolls a fair ordinary dice. He starts with the coin. The probability that he gets a tail on the coin before getting 5 or 6 on the dice, is equal to
Answer & explanation
Correct answer: option 1
Probability of getting 5 or 6 on a specific roll of dice
$\frac{2}{6}=\frac{1}{3}$
and, probability of getting a tail = $\frac{1}{2}$
The desired outcome can happen, in general, on (2r + 1)th trial. That means first 2r trials should result neither in tail nor in 5 or 6, and (2r + 1)th trial must result in tail.
It the conesponding probability is pr then
$p_r=\left(\frac{1}{2}\right)^r . \left(\frac{2}{3}\right)^r . \frac{1}{2}=\frac{1}{2} . \left(\frac{1}{3}\right)^r$
Thus, required probability
$=\sum\limits_{r=0}^{\infty} p_r=\frac{1}{2} \sum\limits_{r=0}^{\infty}\left(\frac{1}{3}\right)^r$
$=\frac{3}{4}$