A person X speaks the truth 4 out of 5 times and person Y speaks the truth 5 out of 6 times. What is the probability that they will contradict each other in stating the fact?

Explanation

Let P(X) = \frac{4}{5} and P(Y) = \frac{5}{6} be the probabilities of speaking the truth. They contradict each other when X speaks the truth and Y lies, or when X lies and Y speaks the truth. This is calculated as P(X)P(Y') + P(X')P(Y) = (\frac{4}{5})(\frac{1}{6}) + (\frac{1}{5})(\frac{5}{6}) = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10}.

Related questions on Statistics & Probability

• Let x be the mean of squares of first n natural numbers and y be the square of mean of first n natural numbers. If $\frac{x}{y}=\fra...
• What is the probability of getting a composite number in the list of natural numbers from 1 to 50?
• Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x+y) is divisible by 4?
• A number x is chosen at random from first n natural numbers. What is the probability that the number chosen satisfies $x+\frac{1}{x} \gt...
• Three fair dice are tossed once. What is the probability that they show different numbers that are in AP?