Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or number on the second die is greater than or equal to the number on the first die?
Solution
When two dice are thrown, the sample space is
∴ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let A be the event that sum of numbers on two dice is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = `("n"("A"))/("n"("S"))= 4/36`
Let B be the event that number on second die is greater than or equal to number on first die.
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)}
∴ n(B) = 21
∴ P(B) = `("n"("B"))/("n"("S")) = 21/36`
Now, (A ∩ B) {(1, 4), (2, 3)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 2/36`
Required Probability is P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= `4/36 + 21/36 - 2/36`
= `23/36`
