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Three zinked coins are given. The probabilities for the head are $ \frac{2}{5} $, $ \frac{3}{5} $ and $ \frac{4}{5} $. A coin is randomly selected and then thrown twice. $ M_k $ denotes the event that the $ k $th coin was chosen with $ k = 1,2,3 $. $ K_j $ stands for the event that at the $j$th throw we get head, where $ j = 1,2 $.

I want to calculate the probability $P(K_2\mid K_1)$.

From the definition of conditional probability we get that $P(K_2\mid K_1)=\frac{P(K_2\cap K_1)}{P(K_1)}$.

Are the events $K_1$ and $K_2$ independent? Does it hold that $P(K_2\cap K_1)=P(K_2)\cdot P( K_1)$ ?

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closed as off-topic by Did, Sahiba Arora, Daniel W. Farlow, NCh, JMP Jul 7 '17 at 4:50

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    $\begingroup$ What is a zinked coin? More to the point: what have you tried? If the first toss is $H$, say, what is the probability that the coin was the first (or second or third) coin? $\endgroup$ – lulu Jul 6 '17 at 17:41
  • $\begingroup$ @lulu, I think that he or she meant unfair coin by zinked. $\endgroup$ – Hasek Jul 6 '17 at 17:43
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    $\begingroup$ Does $H$ at the first throw affects the result of the second throw with any of the three coins? $\endgroup$ – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 17:44
  • $\begingroup$ @Hasek Well, I figured, but I wondered if it was a real phrase. Have you heard it before? A hasty google search didn't turn it up. $\endgroup$ – lulu Jul 6 '17 at 17:52
  • $\begingroup$ @lulu, just to be honest I must admit that also have never heard it. $\endgroup$ – Hasek Jul 6 '17 at 17:57
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The events are dependent because getting H on first toss gives more evidence that $M_3$ happened, making it more likely that I would get second head as well.

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  • $\begingroup$ Ah ok! So, we have to calculate $P(K_2\cap K_1)$, right ? But how? Do we have to make a tree diagram? $\endgroup$ – Mary Star Jul 6 '17 at 18:11
  • $\begingroup$ You can prove dependence by explicitly calculating the probabilities but I find the above argument more elegant. $\endgroup$ – Shadow Jul 6 '17 at 18:20
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Not independent: $$ P(K_1\land K_2)=\frac13\frac25\frac25+\frac13\frac35\frac35+\frac13\frac45\frac45=\frac{29}{75} $$ and $$ P(K_j)=\frac13\frac25+\frac13\frac35+\frac13\frac45=\frac35 $$

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  • $\begingroup$ To calculate $P(K_j)$ why do we multiply each probability by $\frac{1}{3}$ ? $\endgroup$ – Mary Star Jul 6 '17 at 18:30
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    $\begingroup$ Considering that heads come up more than 50% of the time. And, if you have already flipped head, it would suggest that you have a coin more favored to flip heads $P(K_2|K_1)$ should be greater than 50%. $\endgroup$ – Doug M Jul 6 '17 at 18:38
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    $\begingroup$ $P(K_j)$ is the probability that at the $j$th time that we throw the coin we get head. The probability that we have picked the coin $1$ is $\frac{1}{3}$ and its probability to get head is $\frac{2}{5}$. The probability that we have picked the coin $2$ is $\frac{1}{3}$ and its probability to get head is $\frac{3}{5}$. The probability that we have picked the coin $3$ is $\frac{1}{3}$ and its probability to get head is $\frac{4}{5}$. Is this correct? To calculate $P(K_j)$ we take the sum of these, to take into consideration all the possible cases?Do we do the same to calculate $P(K_1\land K_2)$ ? $\endgroup$ – Mary Star Jul 6 '17 at 18:50
  • $\begingroup$ Or is there an other justification? $\endgroup$ – Mary Star Jul 6 '17 at 22:36
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    $\begingroup$ You are correct. If $P(K_j) \neq P(K_1 \cap K_2)$, then the events are not independent. $P(K_j)$ represents the probability that heads is flipped, and you broke this calculation down correctly. The same logic applies for calculating $P(K_1 \cap K_2)$, except heads must be flipped twice in a row, so for each coin you must multiply the probability that it is selected by the probability it turns up heads twice in a row, and take the sum of all possible ways this could happen. $\endgroup$ – TimD1 Jul 7 '17 at 2:47

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