a fair coin is tossed 12 times. the probability of getting at least 8 consecutive heads is 

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Please find below the solution to the asked query:

Let the probability of getting a head be p and that of a tail be q.p=q=12Let X represents the number of times of appearances of a head.So, X can take values 0,1,2,3,4,5,6,7,8,9,10,11 and 12.Pgetting atleast 8 heads=PX8=PX=8+PX=9+PX=10+PX=11+PX=12=C812p8q4+C912p9q3+C1012p10q2+C1112p11q1+C1212p12q0=C812128124+C912129123+C10121210122+C11121211121+C12121212120=C8121212+C9121212+C10121212+C11121212+C12121212=1212C812+C912+C1012+C1112+C1212or C812+C912+C1012+C1112+C12124096=495+220+66+12+14096=7944096=3972048

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