Hint* Probability question

[SnowStar757]

Toss a fair coin until either the number of heads is 3+ the number of tails or the number of tails is 2+ the number of heads (before the first toss the number heads = the number of tails = 0). Let A be the event that at the end, the number of heads is 3 + the number of tails. Find P(A).


Thank you

 

[tkhunny]

Please show us ANYTHING. Almost no one wants to do your homework for you,

 

[SnowStar757]

Actually I figured it out, thanks though

 

[tkhunny]

You're not going to share?

 

[soroban]

Hello, SnowStar757!

I have an approach . . .

\(\displaystyle \text{Toss a fair coin until either: }\) .. \(\displaystyle \begin{Bmatrix}\text{the no. of heads is 3 more than the no. of tails} \\ \text{or} \\ \text{the no. of tails is 2 more than the no. of heads}\end{Bmatrix}\)

\(\displaystyle \text{Let }A\text{ be the event that, at the end, the no. of heads}\) \(\displaystyle \text{is 3 more than the no. of tails.}\)

\(\displaystyle \text{Find }P(A).\)

. . \(\displaystyle \begin{array}{cc}(H,T) & \text{Prob.} \\ \hline \\ (4,1) & {5\choose4}(\frac{1}{2})^5 \\ \\ (5,2) & {7\choose5}(\frac{1}{2})^7 \\ \\ (6,3) & {9\choose6}(\frac{1}{2})^9 \\ \\ (11,7) & {7\choose4}(\frac{1}{2})^{11} \\ \\ \vdots & \vdots \end{array}\)

Therefore: .\(\displaystyle \displaystyle P(A) \;=\;\sum^{\infty}_{n=4} {2n-3\choose n}\left(\frac{1}{2}\right)^{2n-3} \)

Can you evaluate this?