An unbiased die is tossed twice then find the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss ?

CONCEPT:

• Probability of an event is defined as the ratio of number of cases in its favour to the total number of cases
• If A and B are two independent events then P(A ∩ B) = P(A) × P(B)

CALCULATION:

Given: An unbiased die is tossed twice.

In each case, the sample space is given by S = {1, 2, 3, 4, 5, 6}

Let E = event of getting a 4, 5 or 6 on the first toss

Let F =  event of getting a 1, 2, 3 or 4 on the second toss

As we know that, probability of an event is defined as the ratio of number of cases in its favour to the total number of cases

⇒ P(E) = 3/6 = 1/2 and P(F) = 4/6 = 2/3

As we know that, if A and B are two independent events then P(A ∩ B) = P(A) × P(B)

∵ E and F are independent events

⇒ P(E ∩ F) = P(E) × P(F) = (1/2) × (2/3) = 1/3

Hence, the correct option is 4.