February 3rd, 2013, 2:38 am
I'm not getting that answer. I think this is how you solve it:To get exactly two consecutive heads out of n tosses we have:n=2 tosses: {H, H}n=3 tosses: {H, H, T}, {T, H, H}n=4 tosses: {H, H, T, T}, {T, H, H, T}, {T, T, H, H}and so onSimilarly, to get exactly three consecutive heads out of n tosses we have:n=3 tosses: {H, H, H}n=4 tosses: {H, H, H, T}, {T, H, H, H}n=5 tosses: {H, H, H, T, T}, {T, H, H, H, T}, {T, T, H, H, H}and so onAnd for exactly four consecutive heads out of n tosses we haven=4 tosses: {H, H, H, H}n=5 tosses: {T, H, H, H, H}, {H, H, H, H, T}and so onSo to have 2 consecutive heads out of n tosses we have to add all these up2 tosses: {H, H}3 tosses: {H, H, T}, {T, H, H}, {H, H, H}4 tosses: {H, H, T, T}, {T, H, H, T}, {T, T, H, H}, {H, H, H, T}, {T, H, H, H}, {H, H, H, H}so for 2 tosses: 1/2^23 tosses: (2+1)/2^34 tosses: (3+2+1)/2^4n tosses: (n+(n-1)+...+1)/2^n=(n*(n+1))/2^(n+1)