0
$\begingroup$

You flip a weighted coin repeatedly. Assume that, each time the coin is flipped, it lands on heads (H) with probability p > 1/2.

(a) What is the probability that you obtain more heads (H) than tails (T) during the first n flips.

wouldn't the probability just be p>.5 since that is given in the question. I don't know any other way to do this.

$\endgroup$
  • $\begingroup$ If you flip one coin then the probability of more heads than tails is $p$. If you flip, lest's say $100$ coins then the probability is almost $1$ (if $p-\frac12$ is not too small). Can you imagine why? $\endgroup$ – drhab Feb 6 '15 at 10:00
1
$\begingroup$

If n is odd

$$P(\text{more heads than tails}) = \sum_{i = \frac{n+1}{2}}^n {n\choose i} p^i(1-p)^{n-i}$$

If n is even

$$P(\text{more heads than tails}) = \sum_{i = \frac{n+2}{2}}^n {n\choose i} p^i(1-p)^{n-i}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.