Question

A dice is thrown twice. The probability of getting 4, 5 or 6 in the first and 1, 2, 3 or 4 in the second throw is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{3}$

Let P(A) be the probability of getting 4, 5 or 6 in the first throw and P(B) be the probability of getting 1, 2, 3 or 4 in the second throw.
P(A) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
= $\frac{3}{6}$ = $\frac{1}{2}$
P(B) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
= $\frac{4}{6}$ = $\frac{2}{3}$
Now, the probability of getting 4, 5 or 6 in the first and 1, 2, 3 or 4 in the second throw is P(A and B).
$\Rightarrow$ P(A and B) = P(A ∩ B)
We know that P(A ∩ B) = P(A). P(B)
$=\frac{1}{2}\times \frac{2}{3}$
$=\frac{1}{3}$
$\therefore$ The probability of getting 4, 5 or 6 in the first and 1, 2, 3 or 4 in the second throw is $\frac{1}{3}$.