Multiple choice

A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads $90\%$ of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probability that it is the unbiased coin, is:

  1. $\cfrac { 5 }{ 12 } $
  2. $\cfrac { 5 }{ 24 } $
  3. $\cfrac { 1 }{ 3 } $
  4. $\cfrac { 3 }{ 8 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let H be the event of getting heads. The probabilities of picking each coin are 1/3. P(H) = (1/3 * 1) + (1/3 * 0.9) + (1/3 * 0.5) = (1 + 0.9 + 0.5) / 3 = 2.4 / 3 = 0.8. The probability it is the unbiased coin given heads is P(Unbiased and H) / P(H) = (1/3 * 0.5) / 0.8 = 0.5 / 2.4 = 5/24.

AI explanation

Using Bayes theorem, the required probability is the probability of selecting the unbiased coin and getting a head divided by the total probability of getting a head from all three coins. This gives (1/3 multiplied by 1/2) divided by the sum (1/3 multiplied by 1 plus 1/3 multiplied by 0.9 plus 1/3 multiplied by 0.5), which simplifies to 1/6 divided by 8/15. Therefore, the required probability is 5/24.

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