When should we use conditional probability in a coin toss question given x number of tails have already been tossed?

Question

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?

Answer 1

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:

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

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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.

Answer 2

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

$$\frac{\binom{8}{7}}{\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}=\frac{\binom{8}{1}}{\binom{8}{2}+\binom{8}{1}+\binom{8}{0}}=\frac{8}{28+8+1}=\frac{8}{37}$$