A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability 𝑃(ℎ𝑒𝑎𝑑) = 0.5 and for a fake coin, 𝑃(ℎ𝑒𝑎𝑑) = 1. You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)
To solve this problem, apply Bayes' theorem, which helps determine conditional probabilities. Define the events: F as choosing the fake coin, and H as getting two heads when the coin is tossed twice.
We seek P(F|H), the probability that we chose the fake coin given two heads. Bayes' theorem states:
P(F|H) = (P(H|F) * P(F)) / P(H)
Find each component:
- P(F): Probability of picking the fake coin = 1/5.
- P(H|F): Probability of getting two heads with the fake coin (since P(head) = 1) = 1.
- P(H): Total probability of getting two heads across all coins. Use the law of total probability:
P(H)=P(H|F) * P(F) + P(H|R) * P(R)
- P(H|R): Probability of two heads with a regular coin is (0.5)^2 = 0.25.
- P(R): Probability of picking a regular coin = 4/5.
Thus, P(H) = 1 * (1/5) + 0.25 * (4/5) = 1/5 + 1/5 = 2/5.
Substitute these into Bayes' theorem:
P(F|H) = (1 * 1/5) / (2/5) = 1/2 = 0.50
This value, 0.50, falls within the specified range (0.49, 0.51), validating the solution.
Therefore, the probability that the chosen coin is the fake one is 0.50.
Important Questions from Conditional Probability
Two events A and B are such that P(not B) = 0.8, P(A ∪ B) = 0.5 and P(A|B) = 0.4. Then P(A) is equal to
For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?
If an event B has occurred and has P(B) = 1, the conditional probability P(A|B) is equal to:
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
Two integers x and y are chosen with replacement from the set (0, 1, 2…10). The probability that |x - y| > 5 is
