Question

An unbiased dice is tossed twice, Find the probability of getting 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss

1. \(\frac{7}{3}\)
2. \(\frac{18}{23}\)
3. \(\frac{1}{9}\)
4. \(\frac{1}{3}\)

Answer 1

Correct Answer - Option 4 : \(\frac{1}{3}\)

Concept:

If two events are independent then

P(A ∩ B) = P(A) × P(B)

Calculation:

Given:

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

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

Then, P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\), P(F) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Clearly, E and F are independent events.

∴ Required Probability is

P(A ∩ B) = P(A) × P(B)

P(A ∩ B) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\)

P(A ∩ B) = \(\frac{1}{3}\)