JEE MAIN Probability Previous Year Questions (PYQs) – Page 3 of 8
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Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $4m-3n$ is equal to:
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$.$ In a game, a man wins Rs. $100$ if he gets $5$ or $6$ on a throw of a fair die and loses Rs. $50$ for getting any other number. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is:
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Bag $B_1$ contains 6 white and 4 blue balls, Bag $B_2$ contains 4 white and 6 blue balls, and Bag $B_3$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it.
If the ball is white, then the probability that the ball is drawn from Bag $B_2$ is:
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Two different families $A$ and $B$ are blessed with equal number of children. There are $3$ tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family $B$ is $\dfrac{1}{12}$, then the number of children in each family is:
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Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is :
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Solution
Given:
Box I → cards numbered 1 to 30 (30 cards)
Box II → cards numbered 31 to 50 (20 cards)
A box is selected at random → probability of each box = $\dfrac{1}{2}$
Non-prime numbers in each box:
Box I (1–30): Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 → 10 primes.
Non-prime numbers = 30 − 10 = 20
(Including 1 as non-prime)
Box II (31–50): Prime numbers are 31, 37, 41, 43, 47 → 5 primes.
Non-prime numbers = 20 − 5 = 15
Let
A = “card drawn from Box I”
B = “card drawn from Box II”
N = “number on the card is non-prime”
We need $P(A|N)$.
Using Bayes’ theorem:
P(A|N) = \frac{P(A)P(N|A)}{P(A)P(N|A) + P(B)P(N|B)} \]
Now substitute:
\[ P(A) = P(B) = \frac{1}{2}, \quad \] \[ P(N|A) \frac{20}{30} = \frac{2}{3}, \quad \] \[ P(N|B) = \frac{15}{20} = \frac{3}{4} \]
\[ P(A|N) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{3}{4}} \] \[= \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} \] \[= \frac{\frac{1}{3}}{\frac{17}{24}}\] \[= \frac{8}{17} \]
Final Answer: $\boxed{\dfrac{8}{17}}$
Box I → cards numbered 1 to 30 (30 cards)
Box II → cards numbered 31 to 50 (20 cards)
A box is selected at random → probability of each box = $\dfrac{1}{2}$
Non-prime numbers in each box:
Box I (1–30): Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 → 10 primes.
Non-prime numbers = 30 − 10 = 20
(Including 1 as non-prime)
Box II (31–50): Prime numbers are 31, 37, 41, 43, 47 → 5 primes.
Non-prime numbers = 20 − 5 = 15
Let
A = “card drawn from Box I”
B = “card drawn from Box II”
N = “number on the card is non-prime”
We need $P(A|N)$.
Using Bayes’ theorem:
P(A|N) = \frac{P(A)P(N|A)}{P(A)P(N|A) + P(B)P(N|B)} \]
Now substitute:
\[ P(A) = P(B) = \frac{1}{2}, \quad \] \[ P(N|A) \frac{20}{30} = \frac{2}{3}, \quad \] \[ P(N|B) = \frac{15}{20} = \frac{3}{4} \]
\[ P(A|N) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{3}{4}} \] \[= \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} \] \[= \frac{\frac{1}{3}}{\frac{17}{24}}\] \[= \frac{8}{17} \]
Final Answer: $\boxed{\dfrac{8}{17}}$
If three letters can be posted to any one of the $5$ different addresses, then the probability that the three letters are posted to exactly two addresses is:
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Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls.
One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour.
Then the probability that the transferred ball is red is:
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Let A, B and C be three events, which are pair-wise independent and $\overrightarrow E $ denotes the completement of an event E. If $P\left( {A \cap B \cap C} \right) = 0$ and $P\left( C \right) > 0,$ then $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$ is equal to :
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In a class of $60$ students, $40$ opted for NCC, $30$ opted for NSS and $20$ opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is:
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An integer is chosen at random from the integers $1,2,3,\ldots,50$.
The probability that the chosen integer is a multiple of at least one of $4,6$ and $7$ is:
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