Why is this coin-flip not memoryless and independent of past coin-flips for a fair coin, so we have to use bayes theorem and conditional probability?

Question

A man flips a fair coin with sides heads and tails five times. Given that the man receives heads on at least two of the coin flips, what is the probability that he receives tails exactly twice after the five flips of the coin?

A.$\frac{1}{4}$ B.$\frac{5}{16}$ C.$\frac{5}{13}$ D.$\frac{3}{8}$ E.$\frac{8}{13}$

This process is memoryless and we have a $p=\dfrac{1}{2}$

What happened on two flips after five flips is the same as what happens on any two flips.

I think the answer is A but answer key says C.

Answer 1

Given that here should be interpreted as In the condition that, which is a case of Conditional Probability, quoting Wikipedia:

In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred.

We define

• $A$ to be the event that the man receives exactly two tails.
• $B$ to be the event that he gets more than one head.

So the probability of getting at least two heads on 5 flips is

$$ P(B)=1 - \left(\frac 12 \right)^5-\binom{5}{1}\left(\frac 12 \right)^5 = \frac{13}{16} $$

While getting exactly two tails as well as getting at least two heads is

$$ P(A\cap B) = \binom{5}{2}\left(\frac 12 \right)^5 = \frac{5}{16} $$

Note: This value is the same as $P(A)$ because having exactly two tails implies that he's got at least two heads (three, actually).

Using the formula for conditional probability, that is,

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

We get,

$$ P(A\mid B) = \frac{\frac{5}{16}}{\frac{13}{16}} = \frac 5{13} $$

Which is what we wanted.

If $P(A|B) = P(A)$, then $A$ and $B$ must be independent. That is, knowledge about either event does not alter the likelihood of each other. Explained in English: As you already know that there are at least $2$ tails, the probability of it still having $2$ heads is changed relevant to not knowing anything.

Answer 2

A man flips a fair coin with sides heads and tails five times. Given that the man receives heads on at least two of the coin flips, what is the probability that he receives tails exactly twice after the five flips of the coin?

This is poorly phrased.   My reading is that there are no further flips; the question is in regard to the result of these five flips.

To rephrase: "After the five flips: what is the probability that exactly two tails we obtained given that at least two heads were obtained."

Why is this coin-flip not memoryless and independent of past coin-flips for a fair coin, so we have to use bayes theorem and conditional probability?

They are the same five flips, not independent trials.

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So we seek: "The probability for obtaining exactly two tails, given that at most three tails were among the five."

And thus Bayes' Rule is the appropriate to find the answer of $\underline{\phantom{5/13}}$.