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InterviewSolution
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A fair coin is tossed 10 times. Then the probability that two heads donot occur consecutively is`7//64`b. `1//8`c. `9//16`d. `9//64`A. `7//64`B. `1//8`C. `9//16`D. `9//64` |
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Answer» Correct Answer - D Let `p_(i)` denote the probability that out of 10 tosses, head occurs I times and no two heads occur consecutively. If is clear that `igt5.` For I = 0 ,i.e., no head, `p_(0)=1//2^(10).` For I = 1, i.e., ofne heat, `p_(1)=""^(10)C_(1)(1//2)^(1)(1//2)^(9) =10//2^(10).` Now, for i =2, we have 2 heads and 8 tails. Then we have 9 possible places for heads. For example, see the constuctin. `xTxTxTxTxTxTxTxTxTx` Here x represents possible places for heads. Therefore, `P_(2)=""^(9)C_(2)((1)/(2))^(2)(1//2)^(8)=36//2^(10)` Similarly, `P_(3)=""^(8)C_(3)//2^(10)=56//2^(10)` `p_(4)=""^(7)C_(2)//2^(10)=6//2^(10)` `p_(5)=""^(6)C_(5)//2^(10)=6//2^(10)` `thereforep=p_(0)+p_(1)+p_(2)+p_(3)+p_(4)+p_(5)` `=(1+20+36+56+35+6)/(2^(10))=(144)/(2^(10))=9/64` |
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