I'm having a little trouble understanding when we would use conditional probability in the question below.

I was thinking since we are already given that there are at least 6 tails, we wouldn't need to consider that probability, and only calculate the probability of tossing exactly 1 tail (and hence 1 head) in 6th and 7th tosses, which would give 1/2 since the possible outcomes are TH, HT, TT, HH, and 2/4 outcomes have exactly one tail. Could someone please verify if that approach is right or if we'd need to use conditional probability in this case?

Original question:

Assume that the outcome of either heads or tails is equally likely in coin tosses, and each coin toss event occurs independently. You toss the coin exactly 8 times. Given that at least 6 of those tosses resulted in tails, what is the probability that exactly 7 tosses were tails?

  • $\begingroup$ I think you mean "I was thinking since we are already given that there are at least 6 tails" $\endgroup$
    – SagarM
    Oct 29, 2020 at 10:06
  • 1
    $\begingroup$ Yes! Thanks for catching that, edited the question $\endgroup$
    – Jean
    Oct 29, 2020 at 10:11

2 Answers 2


The case you are modelling is that you have tossed 6 tails in a row and now you will toss two more times, and yes then the probability of 7 tails is 1/2.

But the case mentioned in the problem and rightly modelled by @tommik is that someone has tossed 8 coins already (and you cannot see the outcome) and tells you that at least 6 are tails, and then asks you what is the probability that there are exactly 7 tails. Which is modelled as follows:

$$P(\#T=7 | \#T>6) = \frac{P((\#T =7) \& (\#T > 6) )}{P(\#T>6)} = \frac{P(\#T = 7)}{P(\#T>6)}$$

Yeah in retrospect I see that there is slight problem in the way the question is framed, because if you are tossing the coin then it seems to mean that you have seen 6 tails in a row. But I am quite sure what it wants to say is that given such 8 tosses take place and you only have the information that at least 6 are tails then what is the probability of 7 tails.


tossing the coin 8 times, the probability of any single realization is the same for any realization. This because $\mathbb{P}[H]=\mathbb{P}[T]=\frac{1}{2}$

Thus it can be wasted and we are interested only in the combinations.

Thus the answer is



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.