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).
We need to find the conditional probability P(A|B) for rolling an unbiased die once.
The intersection of events A and B (A ∩ B) represents outcomes that are both prime and odd numbers less than 4. From the sets A={2, 3, 5} and B={1, 3}, the common outcome is {3}.
The probability of an event is the number of favorable outcomes divided by the total number of outcomes.
P(A ∩ B) = \frac{|A \cap B|}{|S|} = \frac{1}{6}
P(B) = \frac{|B|}{|S|} = \frac{2}{6} = \frac{1}{3}
The formula for conditional probability is P(A|B) = \frac{P(A \cap B)}{P(B)}.
P(A|B) = \frac{1/6}{1/3}
P(A|B) = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2}
The conditional probability P(A|B) is 0.5.
The question asks for the value rounded off to one decimal place, which is 0.5.
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)