Consider n flips. P(0H)=0.5^n P(1H)=n0.5^n P(2H)=nC2(0.5)^n=n(n-1)/2(0.5)^n. We want the sum of these>0.2
Take out factor 0.5^n
Want (0.5)^n(1+n+n^2/2-n/2)>0.2 That is (0.5)^n(1+n/2+n^2/2)>0.2 By trial and error greatest n=7
This has stumped me.
Assuming a fair coin.
2 flips the P(<=2 H)=1
3 flips the P(<=2 H)=0.875
4 flips the P(<=2 H)=.6875
whats the greatest number of times I can flip a coin and still have P(<=2 H) above 0.2?
Solution thoughts
Break down the probability distribution for 1 to x flips until 0.2 is obtained?
Assume that the data represents a part of a cumulative distribution and find the differrences between the values above sum to one and solve for the unknowns?
Consider n flips. P(0H)=0.5^n P(1H)=n0.5^n P(2H)=nC2(0.5)^n=n(n-1)/2(0.5)^n. We want the sum of these>0.2
Take out factor 0.5^n
Want (0.5)^n(1+n+n^2/2-n/2)>0.2 That is (0.5)^n(1+n/2+n^2/2)>0.2 By trial and error greatest n=7