0
$\begingroup$

The Question

Alice tosses a fair coin seven times. Find the probability that she tosses 4 heads given her first toss is a head. Then, find the probability that she tosses 4 heads given her first and last tosses are heads.

My Work

Part A

$A =$ Alice gets 4 heads $B =$ Her first toss is a head

Part a asks us to find $P(A|B)$

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

We know that $P(B) = \frac{1}{2}$

$P(A\cap B) = P(A)P(B|A)$

$P(A) = \frac{\binom{6}{4}}{2^7}$

$P(B|A) = \frac{P(B\cap A)}{P(A)}$

Now I'm back to where I started trying to find $P(B \cap A)$ can anyone give any hints on how to take a different approach to this problem to get the correct solution?

$\endgroup$
  • $\begingroup$ @VladimirVargas why is that? $\endgroup$ – Dunka Jan 20 '15 at 3:43
  • $\begingroup$ I didn't read "seven times", I'm sorry. $\endgroup$ – Vladimir Vargas Jan 20 '15 at 3:43
  • $\begingroup$ In your calculation (which is not necessary) the probability of $B\cap A$ is $(1/2)\binom{6}{3}(1/2)^6$. $\endgroup$ – André Nicolas Jan 20 '15 at 3:49
3
$\begingroup$

Another way to think about the problem would be "What is the probability of getting 3 heads in the remaining 6 tosses?"

$\endgroup$
  • $\begingroup$ That helped a lot. Thanks $\endgroup$ – Dunka Jan 20 '15 at 4:00

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.