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A bag has 3 coins a, b and c with probability of heads 0.9, 0.6 and 0.5. I flip a randomly drawn coin and get heads. What is the probability that it will come up heads again once re-flipped ?

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  • $\begingroup$ I don't understand how to solve it. Should the proba of what happened (head) be taken into account? $\endgroup$ – astudentofmaths Oct 23 '17 at 16:11
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    $\begingroup$ Yes. This is an exercise in conditional probability. $\endgroup$ – MMASRP63 Oct 23 '17 at 16:14
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probability of coin being 'a' is $\frac{0.9}{0.9+0.6+0.5}$ = 0.45

probability of heads in next turn = 0.9

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probability of coin being 'b' is $\frac{0.6}{0.9+0.6+0.5}$ = 0.3

probability of heads in next turn = 0.6

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probability of coin being 'c' is $\frac{0.5}{0.9+0.6+0.5}$ = 0.25

probability of heads in next turn = 0.5

Total probability = (0.45*0.9)+(0.3*0.6)+(0.25*0.5) = 0.71 \begin{align} Ans = \bbox[yellow,5px,border:2px solid red]{0.71} \end{align}

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  • $\begingroup$ The probability of coin being 'a' is $\frac{1}{3}$. The probability of coin 'a' will be head $0.9$ $\endgroup$ – Hasan Heydari Oct 23 '17 at 16:51
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    $\begingroup$ @HasanHeydari No, your answer is wrong and this one is correct. Having flipped the coin, we have learned information about it. $\endgroup$ – David K Oct 23 '17 at 16:52
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\begin{align} P(\text{Next will be head}|\text{Current is head}) &= \frac{P(\text{Next will be head} \cap \text{Current is head}) }{P(\text{Current is head}) } \\& = \frac{P(\text{Next will be head})P(\text{Current is head})}{P(\text{Current is head})} \\&= P(\text{Next will be head}) \\&= P(a)P(a|a\text{ is head}) + P(b)P(b|b\text{ is head}) + P(c)P(c|c\text{ is head}) \\&= \bbox[yellow,5px,border:2px solid red]{\frac{1}{3}0.9 + \frac{1}{3}0.6 + \frac{1}{3}0.5} \end{align}

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    $\begingroup$ You've just assumed that the probability of flipping heads on the coin you already drew is necessarily equal to the probability of heads if you put the coin back in the bag, drew a new coin randomly, then flipped it. Suppose the bag originally contained one two-headed coin and one two-tailed one, do you still compute this way? $\endgroup$ – David K Oct 23 '17 at 16:52
  • $\begingroup$ @DavidK Current and Next are independent. Thus, my solutions is correct. If the first coin was not returned to the bag, Current and Next are not independent and my solutions is not correct. $\endgroup$ – Hasan Heydari Oct 23 '17 at 16:55
  • $\begingroup$ No, they are not independent, because the problem statement assumes you keep the coin you already had and flip it again. In order to make the flips independent you must repeat the step where you draw a random coin from among the three coins. $\endgroup$ – David K Oct 23 '17 at 16:56
  • $\begingroup$ @HasanHeydari As David noted, your solution is indeed incorrect. The same coin is re-flipped after being drawn. $\endgroup$ – Aaron Montgomery Oct 23 '17 at 18:46

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