Aspire Faculty ID #18685 · Topic: NIMCET 2026 · Just now
NIMCET 2026

A die is rolled twice independently. What is the probability that either the first die shows a number no less than 4 or the second die shows at least 4?

Solution

Let 
$A=$ first die shows a number no less than 4 
So, 
$A={4,5,6}$ 
$P(A)=\frac{3}{6}=\frac{1}{2}$ 
Let $B=$ second die shows at least 4 
So, $B={4,5,6}$ 
$P(B)=\frac{3}{6}=\frac{1}{2}$ 
Required probability: 
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$ 
$=\frac{1}{2}+\frac{1}{2}-\frac{1}{2}\cdot\frac{1}{2}$ 
$=1-\frac{1}{4}$ 
$=\frac{3}{4}$
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