A random variable X has the probability distribution: ![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/60ad1ff9-9c48-4e73-8d5c-c79c - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

hardMCQAIEEE 20042026Mathematics

1 attempts0% success rate3 marks (−1)

A random variable X has the probability distribution: For the events E = {X is a prime number} and F = {X < 4}, the probability P (E \cup F) is

Solution & Step-by-step Explanation

From the distribution:Event E (Prime number X \in {2, 3, 5, 7}): P (E) = P (2) + P (3) + P (5) + P (7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 . Event F (X < 4, so X \in {1, 2, 3}): P (F) = P (1) + P (2) + P (3) = 0.15 + 0.23 + 0.12 = 0.50 . Intersection E \cap F (Prime number and X < 4 \Rightarrow X \in {2, 3}): P (E \cap F) = P (2) + P (3) = 0.23 + 0.12 = 0.35 . P (E \cup F) = P (E) + P (F) - P (E \cap F) = 0.62 + 0.50 - 0.35 = 0.77 .

Try it yourself before checking the explanation above.

A random variable X has the probability distribution: For the events E = {X is a prime number} and F = {X < 4}, the probability P (E \cup F) is

0.87

0.77

0.35

0.50

Practice with timed mock tests and track your performance across Mathematics.