I have a confusing question about unfair coin tosses relating to conditional probability.

Say I have two coins. Coin A has the probability 0.5 of landing on tails. Coin B has the probability 0.6 of landing on tails. What is the probability you will randomly select one of the coins and it will land on tails? And given it does land on tails, what is the conditional probability it was the fair coin?

Any help would be great thanks.

  • $\begingroup$ Just use Bayes theorem. $\endgroup$ – Rohan Feb 22 '17 at 12:57
  • $\begingroup$ its easy for the first but im unsure how to do it for the second $\endgroup$ – Jessie Feb 22 '17 at 12:59

As said in the comments, you just have to use Bayes theorem.

Let $T$ the event "tails", $A$ the event "coin A selected" and $B$ the event "coin B selected". What you're being asked is to find the probability $$P(A\mid T)$$ Using Bayes theorem, we have: $$P(A\mid T) = \frac{P(T\mid A)P(A)}{P(T)}$$

$P(T\mid A)$ is $0.5$ as $A$ is a fair coin. $P(A)$ is $0.5$ (it's not explicitly stated but we can assume that we chose $A$ or $B$ with equal probability). And the denominator is given by the first question.

  • $\begingroup$ Thank you, so would the second part be 1/5? $\endgroup$ – Jessie Feb 22 '17 at 13:10
  • $\begingroup$ I don't think so. What did you find for $P(T)$? $\endgroup$ – Augustin Feb 22 '17 at 13:18
  • $\begingroup$ I have P(A∩Bc) = P(A).(1-P(B)) = (0.5).(1-0.6) = 1/5 $\endgroup$ – Jessie Feb 22 '17 at 14:03
  • $\begingroup$ No the first question asks you to compute $P(T)$, which is $P(T\mid A)P(A) + P(T\mid B)P(B)$. $\endgroup$ – Augustin Feb 22 '17 at 14:29
  • $\begingroup$ Oh i didnt know it had to be certain ways. P(T | A) means the probability of T occuring while A is occuring? $\endgroup$ – Jessie Feb 22 '17 at 15:09

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