A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability π(βπππ) = 0.5 and for a fake coin, π(βπππ) = 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
- A clinic specializes in testing for a disease D. The result of the test can be either positive or negative.
A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.
If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places) - Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let $E_1$ be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let $E_2$ be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.
The conditional probability P($E_1$ | $E_2$) is equal to ____________. (rounded off to one decimal place) - A box contains three apples and two oranges. Two fruits are removed randomly in succession. The probability that the first is an apple and the second is an orange is _____ .
- An experiment is performed by rolling an unbiased die once. Events A and B are defined as follows:
Event A: The outcome of the rolled die is a prime number.
Event B: The outcome of the rolled die is an odd number less than 4.
The conditional probability of A given B, denoted by P(A|B), is ______ (rounded off to one decimal place). - Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is ______ (rounded off to two decimal places).
