Probability of at least one head given that there is at least one tail on three tosses of a fair coin
What is the probability of observing at least one head given that one observes at least one tail on three tosses of a fair coin.
This is a poorly worded question in my opinion and am leaning towards this is a trick question and the answer is simply 1/2. Am I missing any details? This is in a section discussing Bayes Rule and Theorem
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$\begingroup$It would be pretty quick to just list all possible outcomes. There are 8 ways you can flip 3 coins. Only one of them is eliminated by the condition (HHH). How many of the remaining results have at least one head?
$\endgroup$ $\begingroup$This is a poorly worded question. I take it to mean: what is the probability of observing at least one head out of three tosses, given that at least one of these tosses is a tail? Turkeyhundt's answer is nice and intuitive, but I'll do it with the law of conditional probability.
Let $H$ be the number of heads and $T$ the number of tails. Note that $P(H \geq 1) = 1 - P(H=0)$.
Recall the law of conditional probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ So $$P(H \geq 1 | T \geq 1) = \frac{P(H \geq 1 \cap T \geq 1)}{P(T \geq 1)}$$ Which by our earlier observation is equal to: $$P(H \geq 1 | T \geq 1) = \frac{P(H \geq 1 \cap T \geq 1)}{1 - P(T = 0)}$$ Where $P(T = 0)$, by binomial probability and assuming the coin is fair, is $P(T=0) = {3 \choose 0} (1/2)^3(1/2)^0 = 1/8$. Thus: $$P(H \geq 1 | T \geq 1) = \frac{P(H \geq 1 \cap T \geq 1)}{7/8}$$ Now, to find $P(H \geq 1 \cap T \geq 1)$, we should just list the possible sequences of coin tosses that would allow this (there are $8$ tosses). The conditions eliminate two possible sequences: $HHH$ and $TTT$. So $P(H \geq 1 \cap T \geq 1) = 6/8$. Whereby: $$P(H \geq 1 | T \geq 1) = \frac{6/8}{7/8} = \frac{48}{56} = \frac{6}{7}$$
$\endgroup$ $\begingroup$Let $X$ be the random variable counting the number of heads in a toss of 3 coins. So, $X$ follows $B(3,\frac{1}{2})$.
Let $H'$ be the event that at least one of the two remaining tosses produces a heads and $H$ is the event that the first toss is a head. So, we need $P(H'|H)$. $\displaystyle P(H'|H)=\frac{P(H'\cap H)}{P(H)}=\frac{P(X\ge 2)}{P(H)}=\frac{1}{2}$.
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