How many times must a man toss a fair coin, so that the probability of having atleast one head is more than 80%?
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Let the required coin toses be n
Pr(at least one head in n tosses)>0.8
or
Pr(0 heads in n tosses)=<0.2
Probability of 0 heads in n tosses is (1/2)ⁿ
(1/2)ⁿ =< 0.2 take log on both sides
nlog(0.5)=<log(0.2)
n*(-0.3010)=<(-0.6989) or n=>2.3219
∴ approximately it's 3
maheshwaranlupp0yaxy:
Could u explain the first method more clearly
Answered by
0
80 times as the probability of getting coins head is not certain.
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