Question

Two dice each numbered from $1$ to $6$ are thrown together. Let $A$ and $B$ be two events given by
$A:$ even number on the first die
$B:$ number on the second die is greater than $4$

What is $P(A\cup B)$ equal to?

• A. $1/2$
• B. $1/4$
• C. $2/3$
• D. $1/6$

Answer 1

The correct option is C $2/3$

Given:

Two dice are thrown, hence the total number of all possible ways, n(S) = $6\times 6=36$

A: even number on the first die

B: the number on the second die is greater than 4

To find:

$P(A\cup B)$

favourable ways of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Hence n(A)=18

$\therefore ,P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{18}{36}=\frac{1}{2}$

favourable ways of event B = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}

Hence n(B) = 12

$\therefore ,P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{12}{36}=\frac{1}{3}$

Hence, $P(\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)\left(P\right(B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{2}\times \frac{1}{3}=\frac{3+2-1}{6}=\frac{4}{6}=\frac{2}{3}$