Question

If two dice are thrown, what is the probability that at least one of the dice shows a number greater than 3?

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

Answer 1

Answer: (A)

$\frac{3}{4}$

Let A be the event that the first die shows a number greater than 3 and B the event that the second die shows a number greater than 3.
So, $P\left(A\right)=\frac{3}{6}=\frac{1}{2}$
$P\left(B\right)=\frac{3}{6}=\frac{1}{2}$
We have to find at least one of the die shows a number greater than 3 i.e.$P(A\cap B)$
Here, $A\cap B=\left\{\right(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6\left)\right\}$
So, $P(A\cap B)=\frac{9}{36}=\frac{1}{4}$
Now , by addition theorem
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$P(A\cup B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}$

$\Rightarrow P(A\cup B)=\frac{3}{4}$