Two coins are tossed once, where E: tail appears on one coin, F: one coin shows head. The event 'F' refers that it had happened already. The question is what is the probability of finding one tail —the event 'E' happening— from one of two coins if already one coin showing up heads, mathematical statement goes as P(E|F).
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This is often referred to as the Boy-Girl probability paradox. Note that it is not a paradox at all, though people do find it unintuitive. – lulu Feb 09 '22 at 13:59
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1Should add: as is often the case, the "paradox", or the failure of intuition, is linked to the exact phrasing of the problem. You have phrased your question very poorly (we have no idea if "one coin" refers to a specific coin, if it means "at least one coin" or if it means "exactly one coin"). You'll need to clarify those ambiguities before you do he analysis. – lulu Feb 09 '22 at 14:08
1 Answers
In this case it is relevant if you toss two coins at the same time or if you toss the same coin twice.
If you toss the same coin twice the probability of a head in the second try is $\frac 12$ because it is independet from the first try.
But if you have two coins and toss them at the same time and then only look at one coin at first the situation changes. There are four possible outcomes for the two coins: $\{ hh,ht,th,tt\}$ each of them having a probability of $\frac 14$. When we already know that on coin shows head, only three of these possibilities are left, namely $hh,ht,th$. Out of these three (equally probable) possible outcomes two contain atleast one tail - therefore the probability for a tail appearing when we know that one head appeared is $\frac 23$.

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In taking the possible outcomes as {hh,ht,th} since already one coin reflecting itself as head, 'h' aren't the outcomes 'TH' and 'HT' equivalent? – Venkatesa M.G. Feb 09 '22 at 14:14
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When two coins are tossed simultaneously, one of the two lands and shows head. The either coin is still in the air. what is the probability of it facing up tail. – Venkatesa M.G. Feb 09 '22 at 14:21