A fair coin is tossed 10 times. Then the probability that two heads do
not 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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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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