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 on the die is:
A
34
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B
12
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C
13
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D
None
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Solution
The correct option is A34 Probability of getting head = 12 and probability of throwing 5 or 6 with a die =26=13. He starts with a coin and alternately tosses the coin and throws the dice and he will win if he get a head before he get 5 or 6. Hence Probability =12+(12.23).12+(12.23).(12.23).12+.....=12[1+13+(13)2+......]=12.11−13=34