# What is the probabilty that X > 2 conditioning on Y = 2? (Homework)

another homework question here. Let 𝑌 be a binomial random variable with 10 number of trials and 0.2 probability of success. Let X be a uniformly distributed random variable over the interval [0, 3]. If 𝑋 and 𝑌 are independent, what is the probability that 𝑋 > 2 and 𝑌 = 2? What is the probability that 𝑋 > 2 conditioning on 𝑌 = 2?

I already found the probability of X > 2 and Y = 2, which is 5/6 and 0.3020 respectively. Would greatly appreciate hints and thought processes to do this question.

Since the RVs are independent, we have $$P(X>2 \cap Y=2)=P(X>2)P(Y=2)$$ It seems that your $$P(Y=2)={10 \choose 2}(0.2)^2(0.8)^8$$ is correct but $$P(X>2)$$ is not. Share your calculation details for the latter. When it is correct, you’ll just substitute into above equation. The conditional probability is equal to the unconditional one, because the RVs are independent, i.e. $$P(X>2 \vert Y=2) = P(X>2)$$