NIMCET Probability Previous Year Questions (PYQs) – Page 9 of 9
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A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely incontradict each other while narrating some incident is
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Solution
They contradict when one tells the truth and the other lies.
Let \( p_A = 0.6 \) (A tells truth) and \( p_B = 0.5 \) (B tells truth).
\[ P(\text{contradict}) \;=\; P(A\ \text{true},\,B\ \text{lie}) + P(A\ \text{lie},\,B\ \text{true}) \;=\; p_A(1-p_B) + (1-p_A)p_B. \]
\[ =\; 0.6(1-0.5) + (1-0.6)\cdot 0.5 \;=\; 0.6\cdot 0.5 + 0.4\cdot 0.5 \;=\; 0.30 + 0.20 \;=\; 0.50. \]
Answer: \(50\%\).
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?
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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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