A fair coin is tossed $5$ times. It is known that there are more than $2$ heads in the $5$ tosses. What is the probability that there are exactly $3$ heads in the $5$ tosses?

I believe the answer is $\binom{5}{3}$, but I'm confused by the second sentence. How does this change the outcome?

Thanks for answering.

  • 2
    $\begingroup$ Let $H$ be the number of heads in the $5$ tosses. You are looking for $P(H=3|H\ge 2)$. Are you familiar with Bayes rule? $\endgroup$ – mzp Apr 23 '18 at 2:51
  • $\begingroup$ The number of favorable cases is $\binom{5}{3}$, but this cannot be the answer for a probability question since it is larger than $1$. $\endgroup$ – N. F. Taussig Apr 23 '18 at 3:00

$$P(H=3|H> 2)= \frac{P([H=3] \cap [H>2])}{P(H>2)}=\frac{P(H=3)}{P(H>2)}$$

This is a trivial application of Bayes Theorem. What changes the probability here is the fact that we know the number of heads was at least 2. This information reduced the sample space for any further questions asked. For example if we ask what is the probability that number of heads is 1 when it is known that the number of heads was greater than 2. The probability is 0. This is an example of reduction of sample space.

  • $\begingroup$ Thanks for your answer. How were you able to deduce that $$\frac{P([H=3] \cap [H\ge2])}{P(H\ge2)}=\frac{P(H=3)}{P(H\ge2)}?$$ $\endgroup$ – beepboopbeepboop Apr 23 '18 at 3:26
  • $\begingroup$ There are two event that should occur simultaneously. First is the getting 3 heads. Second is getting $\ge 2$ heads. Intersection is the situation when both these occur simultaneously. So how many heads can you have so that both are satisfied?\ $\endgroup$ – Sonal_sqrt Apr 23 '18 at 3:31
  • $\begingroup$ 3 heads. Understood. Thanks! So, P(H=3) = 10/31 and P(H>2) = 16/32. So the final answer would be 5/8, correct? $\endgroup$ – beepboopbeepboop Apr 23 '18 at 3:35
  • $\begingroup$ $P(H\ge 2)=1-P(H=1)-P(H=0)$ check. $\endgroup$ – Sonal_sqrt Apr 23 '18 at 3:37
  • 1
    $\begingroup$ Okay then your answer is correct. $\endgroup$ – Sonal_sqrt Apr 23 '18 at 3:59

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.