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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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