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p=probabilty of getiing 1/3/4 in one throw
q=1-p
P( getting 1,3 or 4 at most n times)=P(0 times) +P(1 time)....P(n times)
=2n+1C0*p0*q2n+1-0
+ 2n+1C1*p1*q2n+1-1 +2n+1C2*p2*q2n+1-2
.....+
2n+1Cn*pn*q2n+1-n
Since, p=q=0.5,
so above = 0.52n+1*[2n+1C0+2n+1C1...2n+1Cn]
= 0.52n+1*[0.5*[2n+1C0+2n+1C1...2n+1C2n+1]]
= 0.52n+1*[0.5*22n+1]
=0.5
hi varsha,
u have a basic method given by akshay sir u can also do dis question in an objective exam by putting n=0 so d question bcomes throwing d die 1 time and we wanna get d probability of getting no 1,3or4 so probability=3/6=0.5 or do by taking n=1 but for subjective do by sir's method
hi varsha ,
the last step was 0.52n+1*[0.5*22n+1].........now , 0.52n+1 =(1/2)2n+1 so 22n+1 within then bracket cancels out with (1/2)2n+1 outside the bracket ,so the answer is 0.5.
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