InterviewSolution
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InterviewSolution
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A fair coin is tossed 100 times. The probability of getting tails 1, 3,.., 49 times is`1//2`b. `1//4`c. `1//8`d. `1//16`A. `1//2`B. `1//4`C. `1//8`D. `1//16` |
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Answer» Correct Answer - B Let the probability of getting a tail in a single trial be `p=1//2.` The number of trials be n=100 and the number of trials in 100 trials be X. `P(X=r)=""^(100)C_(r)p^(r)q^(n-r)` `=""^(100)C_(r)((1)/(2))^(r)((1)/(2))^(100-r)=""^(100)C_(r)((1)/(2))^(100)` Now, `N P(X=1)+P(X=3)+...+P(X=49)` `=""^(100)C_(1)((1)/(2))^(100)+""^(100)C_(3)((1)/(2))^(100)+...+""^(100)C_(49)((1)/(2))^(100)` `But ""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(99)=2^(99)` Also, `""^(100)C_(99)=""^(100)C_(1)` `""^(100)C_(97)=""^(100)C_(3)...""^(100)C_(51)=""^(100)C_(49)` Thus, `2(""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(49)=2^(99)` `or""^(100)C_(1)+""^(100)C_(3)+...+""^(100)C_(49)=2^(98)` Therefore, probability of required event is `(2^(98))/(2^(100))=(1)/(4)2^(98)//2^(100)=1//4` |
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