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The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is : |
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Answer» `p=1/2,q=1-1/2=1/2` `P(r)=nC_r*p^r*q^(n-r)` r success in n trials Required probability=1-P(1)-P(0) `P(1)=nC_1*(1/2)^11*(1/2)^(n-1)` `P(1)=nC_1*(1/2)^n` `P(0)=nC_0*(1/2)^0*(1/2)^n` `=(1/2)^n` `P=1-n*1/(2^n)-1/(2^n)` `=1-1/(2^n)(n+1)>=0.96` `=11-0.96>=1/(2^n)*(n+1)` `0.04>=(n+1)/(2^n)` `2^n/(n+1)>=25` Which is possible when `n>=8` `n=8`. |
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