JEE MAIN Probability Previous Year Questions (PYQs) – Page 4 of 8
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Let $S = \{ M = [a_{ij}], \; a_{ij} \in \{0, 1, 2\}, \; 1 \le i, j \le 2 \}$ be a sample space
and $A = \{ M \in S : M \text{ is invertible} \}$ be an event.
Then $P(A)$ is equal to:
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Solution
Let $\Omega$ be the sample space and $A \subseteq \Omega$ be an event.
Given below are two statements:
(S1): If $P(A)=0$, then $A=\varnothing$
(S2): If $P(A)=1$, then $A=\Omega$
Then:
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Solution
Let $A$ and $B$ be two non-null events such that $A\subset B$. Then, which of the following statements is always correct?
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Solution
If two different numbers are taken from the set {0,1,2,3,...,10} then the probability that their sum as well as absolute difference are both multiple of 4, is :
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Solution
For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = $\dfrac{1}{4}$ and P(All the three events occur simultaneously) =$ \dfrac{1}{16}$. Then the probability that at least one of the events occurs, is :
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Solution
Let the sum of two positive integers be $24$. If the probability that their product is not less than $\dfrac{3}{4}$ times their greatest possible product is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $n-m$ equals
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Solution
If $A$ and $B$ are two events such that $P(A)=0.7,\ P(B)=0.4$ and $P(A\cap \overline{B})=0.5$, where $\overline{B}$ denotes the complement of $B$, then $P!\left(B,\middle|,(A\cup \overline{B})\right)$ is equal to
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Solution
Two integers $x$ and $y$ are chosen with replacement from the set ${0,1,2,3,\ldots,10}$. Then the probability that $|x-y|>5$ is:
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Solution
Let EC denote the complement of an event E.Let E1, E2 and E3 be any pairwise independentevents with P(E1) > 0
and P(E1 $\ \cap $ E2 $ \cap $ E3) = 0.
Then P($E_2^C \cap E_3^C/{E_1}$) is equal to
and P(E1 $\ \cap $ E2 $ \cap $ E3) = 0.
Then P($E_2^C \cap E_3^C/{E_1}$) is equal to
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Solution
Five numbers ${x_1},{x_2},{x_3},{x_4},{x_5}$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $({x_1} < {x_2} < {x_3} < {x_4} < {x_5})$. The probability that ${x_2} = 7$ and ${x_4} = 11$ is :
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Solution
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