Probability
652 Questions
Start JEE Mains Test
2021
Q401
JEE Mains
MCQ
14 Mar 2026
In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :
A.
${{14} \over {45}}$
B.
${{8} \over {45}}$
C.
${{7} \over {45}}$
D.
${{28} \over {45}}$
2021
Q402
JEE Mains
MCQ
14 Mar 2026
When a missile is fired from a ship, the probability that it is intercepted is ${1 \over 3}$ and the probability that the missile hits the target, given that it is not intercepted, is ${3 \over 4}$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is :
A.
${3 \over 4}$
B.
${3 \over 8}$
C.
${1 \over 27}$
D.
${1 \over 8}$
2021
Q403
JEE Mains
MCQ
14 Mar 2026
The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
A.
${1 \over {72}}$
B.
${5 \over {216}}$
C.
${1 \over {36}}$
D.
${1 \over {54}}$
2021
Q404
JEE Mains
MCQ
14 Mar 2026
The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :
A.
${{135} \over {{2^9}}}$
B.
${{65} \over {{2^8}}}$
C.
${{65} \over {{2^7}}}$
D.
${{35} \over {{2^7}}}$
2021
Q405
JEE Mains
MCQ
14 Mar 2026
An ordinary dice is rolled for a certain number of times. If the probability of getting an odd
number 2 times is equal to the probability of getting an even number 3 times, then the
probability of getting an odd number for odd number of times is :
A.
${5 \over {36}}$
B.
${3 \over {16}}$
C.
${1 \over 2}$
D.
${1 \over {32}}$
2021
Q406
JEE Mains
Numerical
14 Mar 2026
Let X be a random variable with distribution.
If the mean of X is 2.3 and variance of X is $\sigma$2, then 100 $\sigma$2 is equal to :
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
If the mean of X is 2.3 and variance of X is $\sigma$2, then 100 $\sigma$2 is equal to :
Correct Answer: 781
Explanation:
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
$\overline X $ = 2.3
$-$a + 6b = ${9 \over {10}}$ ..... (1)
$\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1} $
$a + b = {4 \over {15}}$ .... (2)
From equation (1) and (2)
$a = {1 \over {10}},b = {1 \over 6}$
${\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}} $
${1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}$
$ = {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}$
$ = 4 + a + 3 + 36b - {(2.3)^2}$
$ = 7 + a + 36b - {(2.3)^2}$
$ = 7 + {1 \over {10}} + 6 - {(2.3)^2}$
$ = 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{1310 - {{(23)}^2}} \over {100}}$
$ = {{1310 - 529} \over {100}}$
${\sigma ^2} = {{781} \over {100}}$
$100{\sigma ^2} = 781$
2021
Q407
JEE Mains
Numerical
14 Mar 2026
An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to _____________.
Correct Answer: 28
Explanation:
I1 = first unit is functioning
I2 = second unit is functioning
P(I1) = 0.9, P(I2) = 0.8
P($\overline {{I_1}} $) = 0.1, P($\overline {{I_2}} $) = 0.2
$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {28}}$
$98P = {8 \over {28}} \times 98 = 28$
I2 = second unit is functioning
P(I1) = 0.9, P(I2) = 0.8
P($\overline {{I_1}} $) = 0.1, P($\overline {{I_2}} $) = 0.2
$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {28}}$
$98P = {8 \over {28}} \times 98 = 28$
2021
Q408
JEE Mains
Numerical
14 Mar 2026
The probability distribution of random variable X is given by :
Let p = P(1 < X < 4 | X < 3). If 5p = $\lambda$K, then $\lambda$ equal to ___________.
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X) | K | 2K | 2K | 3K | K |
Let p = P(1 < X < 4 | X < 3). If 5p = $\lambda$K, then $\lambda$ equal to ___________.
Correct Answer: 30
Explanation:
$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$
$ \Rightarrow k = {1 \over 9}$
Now, $p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$
$ \Rightarrow p = {2 \over 3}$
Now, $5p = \lambda k$
$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$
$ \Rightarrow \lambda = 30$
$ \Rightarrow k = {1 \over 9}$
Now, $p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$
$ \Rightarrow p = {2 \over 3}$
Now, $5p = \lambda k$
$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$
$ \Rightarrow \lambda = 30$
2021
Q409
JEE Mains
Numerical
14 Mar 2026
A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ______________.
Correct Answer: 4
Explanation:
P(Head) = ${1 \over 2}$
1 $-$ P(All tail) $\ge$ 0.9
$1 - {\left( {{1 \over 2}} \right)^n}$ $\ge$ 0.9
$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$
$\Rightarrow$ nmin = 4
1 $-$ P(All tail) $\ge$ 0.9
$1 - {\left( {{1 \over 2}} \right)^n}$ $\ge$ 0.9
$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$
$\Rightarrow$ nmin = 4
2021
Q410
JEE Mains
Numerical
14 Mar 2026
Let there be three independent events E1, E2 and E3. The probability that only E1 occurs is $\alpha$, only E2 occurs is $\beta$ and only E3 occurs is $\gamma$. Let 'p' denote the probability of none of events occurs that satisfies the equations
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ and ($\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $ is equal to _____________.
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ and ($\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $ is equal to _____________.
Correct Answer: 6
Explanation:
Let P(E1) = x, P(E2) = y and P(E3) = z
$\alpha $ = P$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$ = $P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$
$ \Rightarrow $ $\alpha $ = x(1 $-$ y) (1 $-$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$\gamma $ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
${x \over {1 - x}} = {\alpha \over p}$
$ \Rightarrow $ x = ${\alpha \over {\alpha + p}}$
From (iii) and (iv)
${z \over {1 - z}} = {\gamma \over p}$
$ \Rightarrow $ z = ${\gamma \over {\gamma + p}}$
$ \therefore $ ${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$ $ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$ ..(v)
Also given,
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ $ \Rightarrow $ $\alpha $p = ($\alpha $ + 2p)$\beta $ ....(vi)
$\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$ $ \Rightarrow $ 3$\gamma $p = (p - 2$\gamma $)$\beta $ .....(vii)
From (vi) and (vii),
${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$
$ \Rightarrow $ p$\alpha $ - 6p$\gamma $ = 5$\gamma $$\alpha $
$ \Rightarrow $ ${p \over \gamma } - {{6p} \over \alpha } = 5$
$ \Rightarrow $ ${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$ ....(viii)
Now from (v) and (viii),
${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$ = 6
$\alpha $ = P$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$ = $P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$
$ \Rightarrow $ $\alpha $ = x(1 $-$ y) (1 $-$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$\gamma $ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
${x \over {1 - x}} = {\alpha \over p}$
$ \Rightarrow $ x = ${\alpha \over {\alpha + p}}$
From (iii) and (iv)
${z \over {1 - z}} = {\gamma \over p}$
$ \Rightarrow $ z = ${\gamma \over {\gamma + p}}$
$ \therefore $ ${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$ $ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$ ..(v)
Also given,
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ $ \Rightarrow $ $\alpha $p = ($\alpha $ + 2p)$\beta $ ....(vi)
$\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$ $ \Rightarrow $ 3$\gamma $p = (p - 2$\gamma $)$\beta $ .....(vii)
From (vi) and (vii),
${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$
$ \Rightarrow $ p$\alpha $ - 6p$\gamma $ = 5$\gamma $$\alpha $
$ \Rightarrow $ ${p \over \gamma } - {{6p} \over \alpha } = 5$
$ \Rightarrow $ ${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$ ....(viii)
Now from (v) and (viii),
${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$ = 6
2021
Q411
JEE Mains
Numerical
14 Mar 2026
Let Bi (i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is $\alpha $, only B2 occurs is $\beta $ and only B3 occurs is $\gamma $. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations $\left( {\alpha - 2\beta } \right)p = \alpha \beta $ and $\left( {\beta - 3\gamma } \right)p = 2\beta \gamma $ (All the probabilities are assumed to lie in the interval (0, 1)).
Then ${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$ is equal to ________.
Then ${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$ is equal to ________.
Correct Answer: 6
Explanation:
Let x, y, z be probability of B1, B2, B3 respectively
$\alpha $ = P(B1 $ \cap $ $\overline {{B_2}} \cap \overline {{B_3}} $) = $P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$
$ \Rightarrow $ x(1 $-$ y)(1 $-$ z) = $\alpha$
Similarly, y(1 $-$ x)(1 $-$ z) = $\beta$
z(1 $-$ x)(1 $-$ y) = $\gamma$
and (1 $-$ x)(1 $-$ y)(1 $-$ z) = p
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$
(x(1 $-$ y)(1 $-$ z) $-$2y(1 $-$ x)(1 $-$ z)) (1 $-$ x)(1 $-$ y)(1 $-$ z) = xy(1 $-$ x)(1 $-$ y)(1 $-$ z)
x $-$ xy $-$ 2y + 2xy = xy
x = 2y ...... (1)
Similarly ($\beta$ $-$ 3$\gamma $)p = 2$\beta$$\gamma $
$ \Rightarrow $ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
${x \over z} = 6$
$\alpha $ = P(B1 $ \cap $ $\overline {{B_2}} \cap \overline {{B_3}} $) = $P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$
$ \Rightarrow $ x(1 $-$ y)(1 $-$ z) = $\alpha$
Similarly, y(1 $-$ x)(1 $-$ z) = $\beta$
z(1 $-$ x)(1 $-$ y) = $\gamma$
and (1 $-$ x)(1 $-$ y)(1 $-$ z) = p
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$
(x(1 $-$ y)(1 $-$ z) $-$2y(1 $-$ x)(1 $-$ z)) (1 $-$ x)(1 $-$ y)(1 $-$ z) = xy(1 $-$ x)(1 $-$ y)(1 $-$ z)
x $-$ xy $-$ 2y + 2xy = xy
x = 2y ...... (1)
Similarly ($\beta$ $-$ 3$\gamma $)p = 2$\beta$$\gamma $
$ \Rightarrow $ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
${x \over z} = 6$
2021
Q412
JEE Advanced
MCQ
14 Mar 2026
Consider three sets E1 = {1, 2, 3}, F1 = {1, 3, 4} and G1 = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E1, and let S1 denote the set of these chosen elements. Let E2 = E1 $-$ S1 and F2 = F1 $\cup$ S1. Now two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements.
Let G2 = G1 $\cup$ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.
Let E3 = E2 $\cup$ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is
Let G2 = G1 $\cup$ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.
Let E3 = E2 $\cup$ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is
A.
${1 \over 5}$
B.
${3 \over 5}$
C.
${1 \over 2}$
D.
${2 \over 5}$
2021
Q413
JEE Advanced
Numerical
14 Mar 2026
A number of chosen at random from the set {1, 2, 3, ....., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is __________.
Correct Answer: 214
Explanation:
Given, set = {1, 2, 3, ...., 2000}
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$\therefore$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
$\therefore$ n(E2) = 285
${E_1} \cap {E_2}$ = multiple of 21 = {21, 42, ....., 1995}
$n({E_1} \cap {E_2}) = 95$
$\therefore$ $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$
$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$ (given)
Hence, $500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$\therefore$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
$\therefore$ n(E2) = 285
${E_1} \cap {E_2}$ = multiple of 21 = {21, 42, ....., 1995}
$n({E_1} \cap {E_2}) = 95$
$\therefore$ $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$
$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$ (given)
Hence, $500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$
2021
Q414
JEE Advanced
Numerical
14 Mar 2026
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of ${{625} \over 4}{p_1}$ is ___________.
The value of ${{625} \over 4}{p_1}$ is ___________.
Correct Answer: 76.25
Explanation:
p1 = 1 $-$ p (all 3 numbers are $\le$ 80)
$ = 1 - {\left( {{{80} \over {100}}} \right)^3} = {{125 - 64} \over {125}} = {{61} \over {125}}$
So, ${{625{p_1}} \over 4} = {{625} \over 4} \times {{61} \over {125}} = {5 \over 4} \times 61 = 76.25$
$ = 1 - {\left( {{{80} \over {100}}} \right)^3} = {{125 - 64} \over {125}} = {{61} \over {125}}$
So, ${{625{p_1}} \over 4} = {{625} \over 4} \times {{61} \over {125}} = {5 \over 4} \times 61 = 76.25$
2021
Q415
JEE Advanced
Numerical
14 Mar 2026
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of ${{125} \over 4}{p_2}$ is ___________.
The value of ${{125} \over 4}{p_2}$ is ___________.
Correct Answer: 24.5
Explanation:
p2 = 1 $-$ p (all three numbers are > 40)
$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$
So, ${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$
$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$
So, ${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$
2021
Q416
JEE Advanced
MSQ
14 Mar 2026
Let E, F and G be three events having probabilities $P(E) = {1 \over 8}$, $P(F) = {1 \over 6}$ and $P(G) = {1 \over 4}$, and let P (E $\cap$ F $\cap$ G) = ${1 \over {10}}$. For any event H, if Hc denotes the complement, then which of the following statements is (are) TRUE?
A.
$P(E \cap F \cap {G^c}) \le {1 \over {40}}$
B.
$P({E^c} \cap F \cap G) \le {1 \over {15}}$
C.
$P(E \cup F \cup G) \le {{13} \over {24}}$
D.
$P({E^c} \cup {F^c} \cup {G^c}) \le {5 \over {12}}$
2021
Q417
AP-EAPCET
MCQ
20 May 2026
A coin is tossed until a head appears or it has been tossed thrice. Given that head doesn’t appear on the first toss, the probability that coin tossed thrice is
A.
$\frac{2}{3}$
B.
$\frac{1}{3}$
C.
$\frac{3}{4}$
D.
$\frac{1}{4}$
2021
Q418
AP-EAPCET
MCQ
20 May 2026
Box-I contains 3 cards bearing numbers 1, 2, 3 , Box II contains 5 cards bearing numbers 1 , 2, 3, 4, 5 and Box III contains 7 cards bearing numbers 1, 2, 3, 4, 5, 6, 7. One card is drawn at random from each of the boxes. If $x_i$ be the number on the card drawn from the $i$ th box, $i=1,2,3$, then the probability that $x_1+x_2+x_3$ is odd is equal to
A.
$\frac{23}{105}$
B.
$\frac{53}{105}$
C.
$\frac{43}{105}$
D.
$\frac{33}{105}$
2021
Q419
AP-EAPCET
MCQ
20 May 2026
The range of a random variable $X$ is $\{1,2,3, \ldots\}$ and $P(X=x)=\frac{C^x}{x !}$. for $x=1,2,3$, ... Then, the value of $C$ is
A.
0
B.
1
C.
ln (2) (where In - denotes the natural log)
D.
$\ln (3)$ (where $\ln$ - denotes the natural log)
2021
Q420
AP-EAPCET
MCQ
20 May 2026
Tom and Jerry play a game of alternately throwing an unfair coin. First one to get head wins. If Tom starts the game, he has 62.5% chance of winning the game. Suppose this coin is tossed 5 times, then the probability of getting exactly 3 head is
A.
$\frac{144}{625}$
B.
$\frac{124}{625}$
C.
$\frac{121}{625}$
D.
$\frac{100}{625}$
2021
Q421
AP-EAPCET
MCQ
20 May 2026
One card is selected at random from 27 cards numbered form 1 to 27. What is the probability that the number on the card is even or divisible by 5.
A.
$\frac{15}{27}$
B.
$\frac{16}{27}$
C.
$\frac{17}{27}$
D.
$\frac{18}{27}$
2021
Q422
AP-EAPCET
MCQ
20 May 2026
Nine balls one drawn simultaneously from a bag containing 5 white and 7 black balls. The probability of drawing 3 white and 6 black balls is
A.
$\frac{{ }^7 C_3}{{ }^{12} C_9}$
B.
$\frac{7}{22}$
C.
$\frac{3}{22}$
D.
$\frac{7}{11}$
2021
Q423
AP-EAPCET
MCQ
20 May 2026
The probabilities that $A$ and $B$ speak truth are $\frac{4}{5}$ and $\frac{3}{4}$ respectively. The probability that they contradict each other when asked to speak on a fact is
A.
$\frac{1}{5}$
B.
$\frac{3}{20}$
C.
$\frac{4}{20}$
D.
$\frac{7}{20}$
2021
Q424
AP-EAPCET
MCQ
20 May 2026
The mean and variance of a binomial variable X are 2 and 1 respectively. The probability that X takes values greater than 1 is
A.
$\frac{5}{16}$
B.
$\frac{8}{16}$
C.
$\frac{11}{16}$
D.
$\frac{1}{16}$
2021
Q425
AP-EAPCET
MCQ
20 May 2026
P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact
A.
38%
B.
48%
C.
52%
D.
62%
2021
Q426
AP-EAPCET
MCQ
20 May 2026
If $A$ and $B$ are two events with $P(A \cap B)=\frac{1}{3}, P(A \cup B)=\frac{5}{6}$ and $P\left(A^C\right)=\frac{1}{2}$, then the value of $P\left(B^C\right)$ is
A.
$\frac{1}{2}$
B.
$\frac{1}{3}$
C.
$\frac{2}{3}$
D.
$\frac{5}{6}$
2021
Q427
AP-EAPCET
MCQ
20 May 2026
A coin is tossed 2020 times. The probability of getting head on 1947th toss is
A.
$\left(\frac{1}{2}\right)^{1947}$
B.
$\left(\frac{1}{2}\right)^{2020}$
C.
$\frac{1}{2}$
D.
$\frac{2}{1947}$
2021
Q428
AP-EAPCET
MCQ
20 May 2026
A discrete random variable X takes values 10, 20, 30 and 40. with probability 0.3, 0.3, 0.2 and 0.2 respectively. Then the expected value of X is
A.
12
B.
22
C.
23
D.
24
2021
Q429
AP-EAPCET
MCQ
20 May 2026
Let $X$ be a random variable which takes values $1,2,3,4$ such that $P(X=r)=K r^3$ where $r=1,2,3,4$ then
A.
$K=\frac{1}{100}$ and $P\left(\frac{1}{2} < X<\frac{5}{2} \mid X > 1\right)=\frac{8}{97}$
B.
$K=\frac{1}{99}$ and $P\left(\frac{1}{2} < X < \frac{5}{2} \mid X > 1\right)=\frac{8}{99}$
C.
$K=\frac{1}{100}$ and $P\left(\frac{1}{2} < X < \frac{5}{2} \mid X > 1\right)=\frac{8}{99}$
D.
$K=\frac{1}{100}$ and $P\left(\frac{1}{2} < X <\frac{5}{2} \mid X >1\right)=\frac{10}{99}$
2021
Q430
AP-EAPCET
MCQ
20 May 2026
12 balls are distributed among 3 boxes, then the probability that the first box will contain 3 balls is
A.
$\frac{{ }^{12} C_3 \times 2^9}{3^{12}}$
B.
$\frac{{ }^{12} C_3 \times 2^9}{3^{10}}$
C.
$\frac{{ }^{12} C_3}{3^{12}}$
D.
$\frac{{ }^{12} C_3}{3^{10}}$
2021
Q431
AP-EAPCET
MCQ
20 May 2026
A random variable X has the probability distribution
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| P(X) | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 |
For the events E = {X is a prime number} and F = {X < 4}, then P(E $\cup$ F) is
A.
0.77
B.
0.87
C.
0.35
D.
0.50
2021
Q432
AP-EAPCET
MCQ
20 May 2026
A die is tossed thrice. If event of getting an even number is a success, then the probability of getting at least 2 successes is
A.
$\frac{7}{8}$
B.
$\frac{1}{4}$
C.
$\frac{2}{3}$
D.
$\frac{1}{2}$
2021
Q433
BITSAT
MCQ
11 Jun 2026
Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings?
A.
${{7} \over {13 }}$
B.
${{63} \over {221 }}$
C.
${{55} \over {221 }}$
D.
${{3} \over {26 }}$
2021
Q434
BITSAT
MCQ
11 Jun 2026
If A and B are two independent events such that $P(A) = {1 \over 2}$ and $P(B) = {1 \over 5}$, then which of the following is correct?
A.
$P\left( {{A \over B}} \right) = {1 \over 2}$
B.
$P\left( {{A \over {A \cup B}}} \right) = {5 \over 6}$
C.
$P\left( {{{A \cap B} \over {A' \cup B'}}} \right) = 0$
D.
All of these
2021
Q435
BITSAT
MCQ
11 Jun 2026
Box I contains 5 red and 2 blue balls, while box II contains 2 red and 6 blue balls. A fair coin is tossed. If it turns up head, a ball is drawn from box I, else a ball is drawn from box II. The probability ball drawn is from box I, if it is blue, is
A.
${{27} \over {56}}$
B.
${{8} \over {29}}$
C.
${{21} \over {29}}$
D.
${{29} \over {56}}$
2020
Q436
JEE Mains
MCQ
14 Mar 2026
The probabilities of three events A, B and C are
given by
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(A$ \cup $B) = 0.8, P(A$ \cap $C) = 0.3, P(A$ \cap $B$ \cap $C) = 0.2, P(B$ \cap $C) = $\beta $
and P(A$ \cup $B$ \cup $C) = $\alpha $, where 0.85 $ \le \alpha \le $ 0.95, then $\beta $ lies in the interval :
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(A$ \cup $B) = 0.8, P(A$ \cap $C) = 0.3, P(A$ \cap $B$ \cap $C) = 0.2, P(B$ \cap $C) = $\beta $
and P(A$ \cup $B$ \cup $C) = $\alpha $, where 0.85 $ \le \alpha \le $ 0.95, then $\beta $ lies in the interval :
A.
[0.35, 0.36]
B.
[0.20, 0.25]
C.
[0.25, 0.35]
D.
[0.36, 0.40]
2020
Q437
JEE Mains
MCQ
14 Mar 2026
Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :
A.
${{10} \over {99}}$
B.
${{5} \over {33}}$
C.
${{15} \over {101}}$
D.
${{5} \over {101}}$
2020
Q438
JEE Mains
MCQ
14 Mar 2026
In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :
A.
${5 \over {6}}$
B.
${5 \over {31}}$
C.
${31 \over {61}}$
D.
${30 \over {61}}$
2020
Q439
JEE Mains
MCQ
14 Mar 2026
The probability that a randomly chosen 5-digit
number is made from exactly two digits is :
A.
${{150} \over {{{10}^4}}}$
B.
${{134} \over {{{10}^4}}}$
C.
${{121} \over {{{10}^4}}}$
D.
${{135} \over {{{10}^4}}}$
2020
Q440
JEE Mains
MCQ
14 Mar 2026
A dice is thrown two times and the sum of the
scores appearing on the die is observed to be
a multiple of 4. Then the conditional probability
that the score 4 has appeared atleast once is :
A.
${1 \over 8}$
B.
${1 \over 9}$
C.
${1 \over 4}$
D.
${1 \over 3}$
2020
Q441
JEE Mains
MCQ
14 Mar 2026
Let EC denote the complement of an event E.
Let E1
, E2
and E3
be any pairwise independent
events 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 :
A.
$P\left( {E_3^C} \right)$ - P(E2)
B.
$P\left( {E_2^C} \right)$ + P(E3)
C.
$P\left( {E_3^C} \right)$ - $P\left( {E_2^C} \right)$
D.
P(E3) - $P\left( {E_2^C} \right)$
2020
Q442
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${8 \over {17}}$
B.
${2 \over 3}$
C.
${2 \over 5}$
D.
${4 \over {17}}$
2020
Q443
JEE Mains
MCQ
14 Mar 2026
If 10 different balls are to be placed in 4 distinct
boxes at random, then the probability that two
of these boxes contain exactly 2 and 3 balls is :
A.
${{965} \over {{2^{11}}}}$
B.
${{965} \over {{2^{10}}}}$
C.
${{945} \over {{2^{11}}}}$
D.
${{945} \over {{2^{10}}}}$
2020
Q444
JEE Mains
MCQ
14 Mar 2026
A random variable X has the following
probability distribution :
Then P(X > 2) is equal to :
| X: | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X): | K2 | 2K | K | 2K | 5K2 |
Then P(X > 2) is equal to :
A.
${1 \over {6}}$
B.
${7 \over {12}}$
C.
${1 \over {36}}$
D.
${23 \over {36}}$
2020
Q445
JEE Mains
MCQ
14 Mar 2026
In a box, there are 20 cards, out of which 10
are lebelled as A and the remaining 10 are
labelled as B. Cards are drawn at random, one
after the other and with replacement, till a
second A-card is obtained. The probability that
the second A-card appears before the third
B-card is :
A.
${{13} \over {16}}$
B.
${{11} \over {16}}$
C.
${{15} \over {16}}$
D.
${{9} \over {16}}$
2020
Q446
JEE Mains
MCQ
14 Mar 2026
Let A and B be two events such that the
probability that exactly one of them occurs is ${2 \over 5}$ and the probability that A or B occurs is ${1 \over 2}$ ,
then the probability of both of them occur
together is :
A.
0.20
B.
0.02
C.
0.01
D.
0.10
2020
Q447
JEE Mains
MCQ
14 Mar 2026
Let A and B be two independent events such
that
P(A) = ${1 \over 3}$ and P(B) = ${1 \over 6}$.
Then, which of the following is TRUE?
P(A) = ${1 \over 3}$ and P(B) = ${1 \over 6}$.
Then, which of the following is TRUE?
A.
$P\left( {{A \over {A \cup B}}} \right) = {1 \over 4}$
B.
$P\left( {{A \over B}} \right) = {2 \over 3}$
C.
$P\left( {{{A'} \over {B'}}} \right) = {1 \over 3}$
D.
$P\left( {{A \over {B'}}} \right) = {1 \over 3}$
2020
Q448
JEE Mains
MCQ
14 Mar 2026
In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ${{1 \over 4}}$
. If the probability that at most two machines will be out of service on the same day is ${\left( {{3 \over 4}} \right)^3}k$, then k is equal to :
A.
${{{17} \over 4}}$
B.
${{{17} \over 2}}$
C.
${{{17} \over 8}}$
D.
4
2020
Q449
JEE Mains
MCQ
14 Mar 2026
An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k
consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected
value of X, is :
A.
$ - {3 \over {16}}$
B.
$ - {1 \over 8}$
C.
${1 \over 8}$
D.
${3 \over {16}}$
2020
Q450
JEE Mains
Numerical
14 Mar 2026
In a bombing attack, there is 50% chance that
a bomb will hit the target. Atleast two
independent hits are required to destroy the
target completely. Then the minimum number
of bombs, that must be dropped to ensure that
there is at least 99% chance of completely
destroying the target, is __________.
Correct Answer: 11
Explanation:
Let n is total no. of bombs being dropped
at least 2 bombs should hit.
P(x > 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - p(x < 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - (p(x = 0) + p(x = 1)) $ \ge $ 0.99
$ \Rightarrow $ 1 - nC0${\left( {{1 \over 2}} \right)^0}{\left( {{1 \over 2}} \right)^n}$ - nC1.${\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{n - 1}}$ $ \ge $ 0.99
$ \Rightarrow $ 1 - ${1 \over {{2^n}}}$ - ${n \over {{2^n}}}$ $ \ge $ ${{99} \over {100}}$
$ \Rightarrow $ ${1 \over {100}}$ $ \ge $ ${{n + 1} \over {{2^n}}}$
$ \Rightarrow $ 2n $ \ge $ 100(n + 1)
Now checking for value of n, we get
n = 11
at least 2 bombs should hit.
P(x > 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - p(x < 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - (p(x = 0) + p(x = 1)) $ \ge $ 0.99
$ \Rightarrow $ 1 - nC0${\left( {{1 \over 2}} \right)^0}{\left( {{1 \over 2}} \right)^n}$ - nC1.${\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{n - 1}}$ $ \ge $ 0.99
$ \Rightarrow $ 1 - ${1 \over {{2^n}}}$ - ${n \over {{2^n}}}$ $ \ge $ ${{99} \over {100}}$
$ \Rightarrow $ ${1 \over {100}}$ $ \ge $ ${{n + 1} \over {{2^n}}}$
$ \Rightarrow $ 2n $ \ge $ 100(n + 1)
Now checking for value of n, we get
n = 11