A man alternately tosses a coin and throws a die 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 `3//4` b. `1//2` c. `1//3` d. none of these

Options

A. `3//4`
B. `1//2`
C. `1//3`
D. None of these

Answer

A

Explanation

The probability of getting a head in a single toss of a coin is P = 1/2 (say). The probability of getting 5 or 6 in a single throw os a die is `q=1//6 =1//3` (say). Therefore, the required probability is
`p+(1-p)(1-q)p+(1-p)(1-q)(1-p)(1-q)p+....`
`=p+(1-p)(1-q)p+(1-p)^(2)(1-q)^(2)p+...`
`=(P)/(1-(1-p)(1-q))`
`=(1//2)/(1-1//2xx2//3)=3/4`