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Standalone Questions
#990
Mathematics
Statistics and Probability
MCQ_SINGLE
APPLY
EASY
2025
JEE Main 2025 (Online) 28th January Evening Shift
KNOWLEDGE
4 Marks
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:
(A) $\frac{2}{5}$
(B) $\frac{4}{15}$
(C) $\frac{1}{3}$
(D) $\frac{2}{3}$
Key: B
Sol:
Sol:
$E_1$: Bag $B_1$ is selected
$B_1$: 6 W 4 B
$B_2$: 4 W 6 B
$B_3$: 5 W 5 B
$E_2$: bag $B_2$ is selected
We have to find $P(\frac{E_2}{A})$
$P(\frac{E_2}{A}) = \frac{P(E_2)P(\frac{A}{E_2})}{P(E_1)P(\frac{A}{E_1}) + P(E_2)P(\frac{A}{E_2}) + P(E_3)P(\frac{A}{E_3})} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} = \frac{4}{15}$
#989
Mathematics
Statistics and Probability
MCQ_SINGLE
APPLY
EASY
2025
JEE Main 2025 (Online) 29th January Evening Shift
KNOWLEDGE
4 Marks
Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $\frac{29}{45}$, then n is equal to:
(A) 5
(B) 6
(C) 4
(D) 3
Key: B
Sol:
Sol:
Bag $1 = \{4 W, 5 B\}$
Bag $2 = \{nW, 3B\}$
$P(\frac{W}{Bag2}) = \frac{29}{45}$
$\Rightarrow P(\frac{W}{B_1}) \times P(\frac{W}{B_2}) + P(\frac{B}{B_1}) \times P(\frac{W}{B_2}) = \frac{29}{45}$
$\frac{4}{9} \times \frac{n+1}{n+4} + \frac{5}{9} \times \frac{n}{n+4} = \frac{29}{45}$
$n = 6$
#988
Mathematics
Statistics and Probability
MCQ_SINGLE
APPLY
EASY
2025
JEE Main 2025 (Online) 2nd April Evening Shift
KNOWLEDGE
4 Marks
Given three indentical bags each containing $10$ balls, whose colours are as follows:
| | Red | Blue | Green |
|--------|-----|------|-------|
| Bag I | $3$ | $2$ | $5$ |
| Bag II | $4$ | $3$ | $3$ |
| Bag III| $5$ | $1$ | $4$ |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $(\frac{1}{p} + \frac{1}{q})$ is:
| | Red | Blue | Green |
|--------|-----|------|-------|
| Bag I | $3$ | $2$ | $5$ |
| Bag II | $4$ | $3$ | $3$ |
| Bag III| $5$ | $1$ | $4$ |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $(\frac{1}{p} + \frac{1}{q})$ is:
(A) $6$
(B) $9$
(C) $7$
(D) $8$
Key: C
Sol:
Sol:
Probability that a Red ball comes from Bag I ($p$):
$p(B_1/R) = \frac{p(B_1) \cdot p(R/B_1)}{p(R)}$
Substituting the values,
$p(B_1/R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$
Simplify to find $p$:
$=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4}$
So, $p = \frac{1}{4}$.
Probability that a Green ball comes from Bag III ($q$):
$p(B_3/G) = \frac{p(B_3) \cdot p(G/B_3)}{p(G)}$
Substitute the values,
$p(B_3/G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}}$
Simplify to find $q$:
$=\frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3}$
So, $q = \frac{1}{3}$.
Calculation of $(\frac{1}{p} + \frac{1}{q})$:
$\frac{1}{p} = 4$, $\frac{1}{q} = 3$
Therefore,
$(\frac{1}{p} + \frac{1}{q}) = 4 + 3 = 7$
#716
Mathematics
Practice
MCQ_SINGLE
APPLY
MEDIUM
2023
KNOWLEDGE
4 Marks
The number of real solutions of the equation x^2 + 5|x| + 6 = 0 is:
Key: 0
Sol:
Sol:
Since |x| is always non-negative, x^2 + 5|x| + 6 is the sum of positive terms plus 6, which can never be zero. Thus, no real solution.
#714
Mathematics
Practice
MCQ_SINGLE
APPLY
MEDIUM
2021
KNOWLEDGE
4 Marks
A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip. The acceleration is:
Key: g tan θ
Sol:
Sol:
For the block to remain stationary relative to the wedge, the pseudo force ma must balance the component of gravity along the incline. ma cosθ = mg sinθ => a = g tanθ.
#713
Mathematics
Practice
MCQ_MULTI
REMEMBER
EASY
2022
KNOWLEDGE
4 Marks
Which of the following ores contain Iron?
Key: Haematite, Magnetite
Sol:
Sol:
Haematite is Fe2O3 and Magnetite is Fe3O4. Bauxite is Al, Malachite is Cu.
#712
Mathematics
Practice
MCQ_SINGLE
APPLY
MEDIUM
2023
KNOWLEDGE
4 Marks
If the complex number z satisfies |z| = 1 and z ≠ -1, then z/(1+z^2) is purely:
Key: Real
Sol:
Sol:
Let z = e^(iθ). Then z/(1+z^2) = e^(iθ) / (1 + e^(2iθ)). Simplifying this using Euler formula results in a real number.
#708
Mathematics
Practice
MCQ_MULTI
REMEMBER
EASY
2022
KNOWLEDGE
4 Marks
Which of the following compounds show Hydrogen bonding?
Key: A, C
Sol:
Sol:
Hydrogen bonding occurs when H is bonded to highly electronegative atoms like F, O, or N.
#707
Mathematics
Practice
MCQ_SINGLE
APPLY
MEDIUM
2023
KNOWLEDGE
4 Marks
A particle moves along a straight line such that its displacement x changes with time t as x = t^3 - 6t^2 + 3t + 4. What is the velocity when acceleration is zero?
Key: -9 m/s
Sol:
Sol:
Velocity v = dx/dt = 3t^2 - 12t + 3. Acceleration a = dv/dt = 6t - 12. For a=0, 6t=12 => t=2s. At t=2, v = 3(4) - 12(2) + 3 = -9.
#705
Mathematics
Statistics and Probability