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Two coins are in a bag. One is symmetric, while the other is not - if tossed it lands heads up with probability equal to 0.6. One coin is randomly pulled out of the bad and tossed. It lands heads up. What is the probability that the same coin will land heads up if tossed again?

My attempt is we denote

$C_1$ = {the symmetric coin is chosen}

$C_2$ = {the nonsymmetric coin is chosen}

$H_1$ = {1st toss is head}

$H_2$ = {2nd toss is head}

Then it seems $$P(H_2 | H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)}$$

is the answer? But I think this is wrong. The question requires the same coin.

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  • $\begingroup$ Please see this article on MathSE protocol. As onerous as the article may appear to you, it provides a defense mechanism against the MathSE forum being used as a do my homework forum. In particular, please see the Edit-Tools section of the article, and the portion of the article that discusses showing work. $\endgroup$ Nov 4, 2022 at 21:39
  • $\begingroup$ You haven't actually shown work, with numbers. Also, you don't explain how you are going to use $C_1$ or $C_2$. See Bayes Theorem. $\endgroup$ Nov 4, 2022 at 21:42
  • $\begingroup$ Does a "symmetric" coin mean that both faces are the same? $\endgroup$
    – peterwhy
    Nov 4, 2022 at 23:10
  • $\begingroup$ @peterwhy yes. symmetric means P(head)=0.5 $\endgroup$
    – Jonathen
    Nov 4, 2022 at 23:21

1 Answer 1

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By the process of marginalization,

$P(H_2|H1) = P(H2, C1|H1) + P(H2, C2|H1) = P(H2|C1, H1)P(C1|H1) + P(H2|C2, H1)P(C2|H1) = 0.5P(C1|H1) + 0.6P(C2|H1)$.

Now use Bayes' rule to calculate each of these probabilities:

$P(C1|H1) = \frac{P(H1|C1)P(C1)}{P(H1)} = \frac{0.5 \times 0.5}{0.5 \times 0.5 + 0.6 \times 0.5} = 5/11$

$P(C2|H1) = 6/11$

So $P(H2|H1) = (2.5 + 3.6)/11 = 61/110$

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  • $\begingroup$ My concern is $P(H_2 | H_1)$, this value or the event, doesn't count on the requirement that the 2nd coin is the same as the first time tossed coin. $\endgroup$
    – Jonathen
    Nov 4, 2022 at 23:24
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    $\begingroup$ It does. There is no second draw. C1 and C2 here indicate the coin that is chosen initially before H1. Otherwise this would be essentially finding $P(H1)$ by independence. $\endgroup$ Nov 4, 2022 at 23:26

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