JEE Main Previous Year Probability Questions and Solutions | Addition Theorem, Conditional Probability and Bayes’ Theorem

Q 21 :

The probability that the birthdays of six different persons will fall in exactly two calendar months

$\frac{342}{12^{5}}$
$\frac{341}{12^{6}}$
$\frac{341}{12^{5}}$
$\frac{342}{12^{6}}$

1

2

3

4

(3)

$n (S) = 12^{6}$

Now, any two months can be chosen in ${}^{12}C_{2}$ ways. The six birthdays can fall in these two months in $2^{6}$ ways. Out of these $2^{6}$ ways there are two ways when all the six birthdays fall in one month so favourable number of ways is ${}^{12}C_{2} \times (2^{6} - 2)$ .

Hence required probability is

$= \frac{{}^{12}C_{2} \times (2^{6} - 2)}{12^{6}}$

$= \frac{12 \times 11 \times (2^{6} - 2)}{2 \times 12^{6}}$

$= \frac{341}{12^{5}}$

Q 22 :

Two dice are thrown simultaneously to get the co-ordinates on $x - y$ plane. Then the probability that this point lies inside or on the region bounded by $| x |$ $+$ $| y |$ $= 3$ , is:

$\frac{3}{14}$
$\frac{2}{3}$
$\frac{1}{12}$
$\frac{4}{14}$

1

2

3

4

(3)

$favorable points \to (1, 1) (1, 2) (2, 1)$

$Prob. = \frac{3}{26} = \frac{1}{12}$

Q 23 :

Number of correct statements among the following is $k$ . Then $3 k$ is ________

Statement I: For two given events A and B, $P (A \cap B)$ is not less than $P (A) + P (B) - 1$ .

Statement II: The number of symmetric relations defined on the set {1,2,3,4} which are not reflexive is 20.

Statement III: $\lim_{n \to \infty} {(a^{\frac{1}{^{2}}} - a^{\frac{1}{^{3}}}) (a^{\frac{1}{^{2}}} - a^{\frac{1}{^{5}}}) \dots (a^{\frac{1}{^{2}}} - a^{\frac{1}{^{2 n + 1}}})} = 0,$ $if a > 1 .$

(6)

$Statement-I: Use addition theorem of probability$

$Statement-II: Total number of relation both symmetric and reflexive = 2^{(\frac{n^{2} - n}{2})}$

$And total number of symmetric relation = 2^{(\frac{n^{2} + n}{2})}$

$Then number of symmetric relation which are not reflexive = 2^{\frac{n (n + 1)}{2}} - 2^{\frac{n (n - 1)}{2}}$

$= 2^{10} - 2^{6} = 1024 - 64 = 960$

Q 24 :

Nine balls of the same size and colour, numbered $1, 2, \dots, 9$ , were put into an Urn. Now A draws a ball from Urn, noted that it is of number $x$ , and puts it back. Then B also drawn a ball from the Urn and noted that it is of number $y$ . Then probability that the inequality $x - 2 y + 10 > 0$ to hold is

$\frac{52}{18}$
$\frac{59}{81}$
$\frac{60}{81}$
$\frac{61}{81}$

1

2

3

4

(4)

Since each has equally 9 different possible results for A and B to draw a ball from the packet independently, the total number of possible events is $9^{2} = 81$ .

From $a - 2 b + 10 > 0$ , we get $2 b < a + 10$ . We find that when $b$ = 1, 2, 3, 4, 5, a can take any value from 1, 2, 3, ..., 9 to make the inequality hold. Then we have 9 × 5 = 45 admissible events.

When b = 6, a can be 3, 4, ..., 9 and there are 7 admissible events.

When b = 7, a can be 5, 6, 7, 8, 9 and there are 5 admissible events.

When b = 8, a can be 7, 8, 9 and there are 3 admissible events.

When b = 9, a can be 9 and there is 1 admissible event.

So, the required probability is $\frac{45 + 7 + 5 + 3 + 1}{81} = \frac{61}{81}$

Q 25 :

Let N denotes the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to

(8)

$2^{N} < N! Which is true when N \geq 4$

$N = 1 (Not possible)$

$N = 2 i.e., (1, 1) (Not possible)$

$∴$ $required probability = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$

$∴$ $m = 11 and n = 12$

$Now, 4 m - 3 n = 4 (11) - 3 (12) = 44 - 36 = 8$

Q 26 :

Let there be three independent events $E_{1}, E_{2}$ and $E_{3}$ . The probability that only $E_{1}$ occurs is $α$ , only $E_{2}$ occurs is $β$ and only $E_{3}$ occurs is $γ$ . Let $' p'$ denote the probability of none of events occurs that satisfies the equation $(α - 2 β) p = α β$ and $(β - 3 γ) p = 2 β γ$ . 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

9
3
7
6

1

2

3

4

(4)

$Let p (E_{1}) = x, p (E_{2}) = y and p (E_{3}) = z$

$α = p (E_{1} \cap \bar{E_{2}} \cap \bar{E_{3}}) = p (E_{1}) \cdot p (\bar{E_{2}}) \cdot p (\bar{E_{3}})$

$\Rightarrow α = x (1 - y) (1 - z)$ $\dots \dots (i)$

$Similarly,$

$β = (1 - x) y (1 - z)$ $\dots \dots (i i)$

$γ = (1 - x) (1 - y) z$ $\dots \dots (i i i)$

$p = (1 - x) (1 - y) (1 - z)$ $\dots \dots (i v)$ $and solve equation.$

Q 27 :

An urn contains 7 white and 5 black balls. A ball is drawn at random and is put back into the urn along with 3 additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Then the probability that the ball drawn is white is

$\frac{7}{32}$
$\frac{5}{12}$
$\frac{7}{12}$
$\frac{10}{25}$

1

2

3

4

(3)

$W_{1} = ball drawn in the first draw is white$

$W_{2} = ball drawn in the second draw is white,$ $P (W_{1}) = \frac{7}{12}$

$P (B_{1}) = \frac{5}{12},$ $P (W_{2} / W_{1}) = \frac{10}{15},$ $P (W_{2} / B_{1}) = \frac{7}{15}$

$P (W_{2}) = P (W_{1}) \times P (W_{2} / W_{1}) + P (B_{1}) \times P (W_{2} / B_{1})$

Q 28 :

An urn contains 7 green and 5 yellow balls. Two balls are drawn at a time. The probability that both balls are of the same colour is

$\frac{1}{33}$
$\frac{5}{33}$
$\frac{7}{22}$
$\frac{31}{66}$

1

2

3

4

(4)

$Reqd. prob. = \frac{{}^{7}C_{2}}{{}^{12}C_{2}} + \frac{{}^{5}C_{2}}{{}^{12}C_{2}} = \frac{21 + 10}{66} = \frac{31}{66}$

Q 29 :

Let two fair six-faced dice A and B be thrown simultaneously. If $E_{1}$ is the event that die A shows up four, $E_{2}$ is the event that die B shows up two, then which of the following statements is true?

$E_{1}$ and $E_{2}$ are exclusive
$E_{2}$ and $E_{1}$ are exhaustive
$E_{1}$ and $E_{2}$ are dependent
$E_{1}$ , $E_{2}$ are independent

1

2

3

4

(4)

$E_{1} = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}$

$E_{2} = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}$

$E_{3} = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}$

$P (E_{1}) = \frac{1}{6}; P (E_{2}) = \frac{1}{6}; P (E_{3}) = \frac{1}{2}$

$P (E_{1} \cap E_{2}) = \frac{1}{36}, P (E_{2} \cap E_{3}) = \frac{1}{12}, P (E_{1} \cap E_{3}) = \frac{1}{12}$

And $P (E_{1} \cap E_{2} \cap E_{3}) = 0 \neq P (E_{1}) . P (E_{2}) . P (E_{3})$

$\Rightarrow 0 \neq \frac{1}{6} . \frac{1}{6} . \frac{1}{2} \Rightarrow E_{1}, E_{2}, E_{3} are not independent.$

Q 30 :

Two dice are rolled. If the probability of getting different numbers on these dice is $\frac{p}{q}$ where $p, q$ are relatively prime then $q - p$ is ____

4
3
1
2

1

2

3

4

(3)

$P (E) = \frac{6 \times 5}{6 \times 6} = \frac{5}{6}$

Q 31 :

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$ , $g c d (m, n) = 1$ , then $m + n$ is equal to: [2026]

53
55
107
105

1

2

3

4

(3)

A head gives 10 points, a tail gives 5 points.

For score to be 30,

HHH i.e., 30 pts = 1 way

TTTTTT i.e., 30 pts = 1 way

HHTT i.e., $\frac{4!}{2! 2!} = 6$ ways

TTTTH i.e., $\frac{5!}{4!} = 5$ ways

Required probability

$= \frac{1}{2^{3}} + \frac{1}{2^{6}} + \frac{6}{2^{4}} + \frac{5}{2^{5}} = \frac{1}{8} + \frac{1}{64} + \frac{6}{16} + \frac{5}{32}$

$= \frac{8 + 1 + 24 + 10}{64} = \frac{43}{64} = \frac{m}{n}$ $(∵ Given)$

$∴$ $m = 43, n = 64$

So, $m + n = 107$

Q 32 :

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is: [2026]

$\frac{7}{10}$
$\frac{10}{17}$
$\frac{12}{19}$
$\frac{7}{19}$

1

2

3

4

(2)

Let $E_{1}$ and $E_{2}$ be the events that the letter is from KANPUR and ANANTPUR respectively.

Let A be the event that letters 'AN' are visible on the envelope.

$P (E_{1}) = \frac{1}{2} = P (E_{2})$

The pairs in KANPUR are ${KA, AN, NP, PU, UR}$ i.e., 5 in number.

The pairs in ANANTPUR are ${AN, NA, AN, NT, TP, PU, UR}$ i.e., 7 in number.

So, $P (A | E_{1}) = \frac{1}{5}$ and $P (A | E_{2}) = \frac{2}{7}$

$∴$ $P (\frac{E_{2}}{A}) = \frac{P (E_{1}) P (\frac{A}{E_{1}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}})}$

$= \frac{\frac{1}{2} \times \frac{2}{7}}{\frac{1}{2} \times \frac{1}{5} + \frac{1}{2} \times \frac{2}{7}} = \frac{10}{17}$

Q 33 :

The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is : [2026]

0.74
0.76
0.72
0.78

1

2

3

4

(2)

Given, $P (A selected) = 0.6$ and $P (B selected) = 0.4$

Also, $P (win | A) = 0.8$ , $P (win | B) = 0.7$

$∴$ $P (win) = P (A) \cdot P (\frac{win}{A}) + P (B) \cdot P (\frac{win}{B})$

$= (0.6) (0.8) + (0.4) (0.7) = 0.48 + 0.28$

$= 0.76$

Q 34 :

A bag contains $(N + 1)$ coins - $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$ , then $N$ is equal to : [2026]

5
7
8
9

1

2

3

4

(2)

A bag has $(N + 1)$ coins i.e., $N$ fair coins and 1 coin with 2 heads

Probability of getting head

$= \frac{N}{N + 1} \times \frac{1}{2} + \frac{1}{N + 1} \times 1$

$= \frac{N}{2 (N + 1)} + \frac{1}{N + 1} = \frac{N + 2}{2 (N + 1)}$

$Also, \frac{N + 2}{2 (N + 1)} = \frac{9}{16}$ $(Given)$

$\Rightarrow 16 (\frac{N + 2}{2}) = 9 (N + 1)$

$\Rightarrow 8 N + 16 = 9 N + 9 \Rightarrow N = 7$

Q 35 :

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is : [2026]

$\frac{63}{925}$
$\frac{17}{231}$
$\frac{16}{231}$
$\frac{64}{925}$

1

2

3

4

(3)

Required probability

$= \frac{{}^{6}{C_{1}} \times {}^{6}{C_{1}}}{{}^{12}{C_{2}}} \times \frac{{}^{5}{C_{1}} \times {}^{5}{C_{1}}}{{}^{10}{C_{2}}} \times \frac{{}^{4}{C_{1}} \times {}^{4}{C_{1}}}{{}^{8}{C_{2}}}$ $\times \frac{{}^{3}{C_{1}} \times {}^{3}{C_{1}}}{{}^{6}{C_{2}}} \times \frac{{}^{2}{C_{1}} \times {}^{2}{C_{1}}}{{}^{4}{C_{2}}} \times \frac{{}^{1}{C_{1}} \times {}^{1}{C_{1}}}{{}^{2}{C_{2}}}$

$= \frac{36}{66} \times \frac{25}{45} \times \frac{16}{28} \times \frac{9}{15} \times \frac{4}{6} \times \frac{1}{1} = \frac{16}{231}$

Q 36 :

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5}, \frac{1}{5}$ and $\frac{2}{5}$ . The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}, \frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is : [2026]

$\frac{11}{37}$
$\frac{12}{37}$
$\frac{13}{37}$
$\frac{14}{37}$

1

2

3

4

(2)

Let $E_{1}, E_{2}, E_{3}$ be the events that candidate travel by bus, scooter and car respectively.

$P (E_{1}) = \frac{2}{5}, P (E_{2}) = \frac{1}{5}, P (E_{3}) = \frac{2}{5}$

Let 'A' be the event that candidate reaches late.

$∴$ $P (\frac{A}{E_{1}}) = \frac{1}{5}, P (\frac{A}{E_{2}}) = \frac{1}{3}, P (\frac{A}{E_{3}}) = \frac{1}{4}$

Now,

$P (\frac{E_{1}}{A}) = \frac{P (E_{1}) \cdot P (\frac{A}{E_{1}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) \cdot P (\frac{A}{E_{2}}) + P (E_{3}) \cdot (\frac{A}{E_{3}})}$

$= \frac{\frac{2}{5} \times \frac{1}{5}}{\frac{2}{5} \times \frac{1}{5} + \frac{1}{5} \times \frac{1}{3} + \frac{2}{5} \times \frac{1}{4}} = \frac{12}{37}$

Q 37 :

Let $a, b, c \in {1, 2, 3, 4}$ . If the probability, that $a x^{2} + 2 \sqrt{2} b x + c > 0$ for all $x \in R$ , is $\frac{m}{n}$ , $\gcd (m, n) = 1$ , then $m + n$ is equal to ________. [2026]

(81)

$D < 0 & a > 0$

$\Rightarrow$ ${(2 \sqrt{2})}^{2} b^{2} - 4 a c < 0 \Rightarrow 8 b^{2} - 4 a c < 0$

$\Rightarrow$ $2 b^{2} < a c$

When $b = 1$ , $a c > 2$

then $(a, c)$ = (1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (4, 4) i.e., 13 pairs

When $b = 2$ , $a c > 8$

then $(a, c)$ = (3, 3), (3, 4), (4, 3), (4, 4) i.e., 4 pairs

when $b = 3$ , $a c > 18$ , which is impossible

when $b = 4$ , $a c > 32$ , which is impossible

$∴$ Favourable cases = 13 + 4 = 17

Total cases $= 4^{3} = 64$

$∴$ Required probability $= \frac{17}{64} = \frac{m}{n}$ (Given)

So, $m + n = 81$

Q 38 :

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $p$ , then $96 p$ is equal to ________. [2026]

(9)

We have, $n = 8$

$P (x = 4 in the first 6 tosses)$ $= {}^{6}{C_{4}} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{2}$

$P (One head at 4^{th}, 5^{th} or 6^{th} toss)$

$= {}^{3}{C_{1}} (\frac{1}{2}) {(\frac{1}{2})}^{2} \times \frac{1}{2} \times \frac{1}{2^{3}} = \frac{3}{2^{8}}$

$P (Two head at 4^{th}, 5^{th} or 6^{th} toss)$

$= {}^{3}{C_{2}} {(\frac{1}{2})}^{2} (\frac{1}{2}) \times {}^{3}{C_{2}} {(\frac{1}{2})}^{2} \times \frac{1}{2} \times {}^{2}{C_{1}} (\frac{1}{2}) (\frac{1}{2})$

$= \frac{3 \times 3 \times 2}{2^{8}} = \frac{18}{2^{8}}$

$P (Three head at 4^{th}, 5^{th} & 6^{th} toss)$

$= {}^{3}{C_{3}} {(\frac{1}{2})}^{3} \times {}^{3}{C_{1}} (\frac{1}{2}) \cdot {(\frac{1}{2})}^{2} \times {(\frac{1}{2})}^{2} = \frac{3}{2^{8}}$

Now, $p = \frac{3 + 18 + 3}{2^{8}} = \frac{24}{2^{8}} = \frac{4 \times 2 \times 3}{2^{8}} = \frac{3}{32}$

Thus, $96 p = 96 \times \frac{3}{32} = 9$

Q 39 :

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$ , where $a, b \in N$ and $g c d (a, b) = 1$ , then $a + b$ is equal to ________. [2026]

(944)

$n (s) = {}^{31}{C_{3}} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$

'3' selected dates are increasing order in A.P.

$x, x + d, x + 2 d d \geq 1$ (common difference)

Here, $x \geq 1 & x + 2 d \leq 31$

If $d = 1$ , then $x \leq 29 \Rightarrow 29$ choices for $x$

If $d = 2$ , then $x \leq 27 \Rightarrow 27$ choices for $x$

$⋮$

If $d = 15,$ $x \leq 1 \Rightarrow 1$ choice for $x$

Total favourable cases,

$n (E) = 1 + 3 + 5 + \dots + 29$

$= \frac{15}{2} (29 + 1) = 225$

So, required probability $= \frac{225}{4495} = \frac{45}{899} = \frac{a}{b}$ $[∵ Given]$

$\Rightarrow$ $a = 45, b = 899$

So, $a + b = 45 + 899 = 944$

Topic Question Set

Q 21 :

The probability that the birthdays of six different persons will fall in exactly two calendar months

Q 22 :

Two dice are thrown simultaneously to get the co-ordinates on $x - y$ plane. Then the probability that this point lies inside or on the region bounded by $| x |$ $+$ $| y |$ $= 3$ , is:

Q 25 :

Let N denotes the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to

Q 27 :

An urn contains 7 white and 5 black balls. A ball is drawn at random and is put back into the urn along with 3 additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Then the probability that the ball drawn is white is

Q 28 :

An urn contains 7 green and 5 yellow balls. Two balls are drawn at a time. The probability that both balls are of the same colour is

Q 29 :

Let two fair six-faced dice A and B be thrown simultaneously. If $E_{1}$ is the event that die A shows up four, $E_{2}$ is the event that die B shows up two, then which of the following statements is true?

Q 30 :

Two dice are rolled. If the probability of getting different numbers on these dice is $\frac{p}{q}$ where $p, q$ are relatively prime then $q - p$ is ____

Q 31 :

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$ , $g c d (m, n) = 1$ , then $m + n$ is equal to: [2026]

Q 32 :

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is: [2026]

Q 34 :

A bag contains $(N + 1)$ coins - $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$ , then $N$ is equal to : [2026]

Q 35 :

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is : [2026]

Q 37 :

Let $a, b, c \in {1, 2, 3, 4}$ . If the probability, that $a x^{2} + 2 \sqrt{2} b x + c > 0$ for all $x \in R$ , is $\frac{m}{n}$ , $\gcd (m, n) = 1$ , then $m + n$ is equal to ________. [2026]

Q 38 :

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $p$ , then $96 p$ is equal to ________. [2026]

Q 39 :

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$ , where $a, b \in N$ and $g c d (a, b) = 1$ , then $a + b$ is equal to ________. [2026]

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Topic Question Set

Q 21 : The probability that the birthdays of six different persons will fall in exactly two calendar months

Q 22 : Two dice are thrown simultaneously to get the co-ordinates on x-y plane. Then the probability that this point lies inside or on the region bounded by |x|+|y|=3, is:

Q 24 : Nine balls of the same size and colour, numbered 1,2,…,9, were put into an Urn. Now A draws a ball from Urn, noted that it is of number x, and puts it back. Then B also drawn a ball from the Urn and noted that it is of number y. Then probability that the inequality x-2y+10>0 to hold is

Q 25 : Let N denotes the sum of the numbers obtained when two dice are rolled. If the probability that 2N<N! is mn, where m and n are coprime, then 4m-3n is equal to

Q 27 : An urn contains 7 white and 5 black balls. A ball is drawn at random and is put back into the urn along with 3 additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Then the probability that the ball drawn is white is

Q 28 : An urn contains 7 green and 5 yellow balls. Two balls are drawn at a time. The probability that both balls are of the same colour is

Q 29 : Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two, then which of the following statements is true?

Q 30 : Two dice are rolled. If the probability of getting different numbers on these dice is pq where p,q are relatively prime then q-p is ____

Q 31 : A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is mn, gcd(m, n)=1, then m+n is equal to: [2026]

Q 32 : A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is: [2026]

Q 34 : A bag contains (N+1) coins -N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is 916, then N is equal to : [2026]

Q 35 : A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is : [2026]

Q 37 : Let a,b,c∈{1,2,3,4}. If the probability, that ax2+22bx+c>0 for all x∈R, is mn, gcd(m,n)=1, then m+n is equal to ________. [2026]

Q 38 : A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is p, then 96p is equal to ________. [2026]

Q 39 : From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ab, where a,b∈N and gcd (a, b)=1, then a+b is equal to ________. [2026]

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Q 21 :

The probability that the birthdays of six different persons will fall in exactly two calendar months

Q 22 :

Two dice are thrown simultaneously to get the co-ordinates on $x - y$ plane. Then the probability that this point lies inside or on the region bounded by $| x |$ $+$ $| y |$ $= 3$ , is:

Q 25 :

Let N denotes the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to

Q 27 :

An urn contains 7 white and 5 black balls. A ball is drawn at random and is put back into the urn along with 3 additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Then the probability that the ball drawn is white is

Q 28 :

An urn contains 7 green and 5 yellow balls. Two balls are drawn at a time. The probability that both balls are of the same colour is

Q 29 :

Let two fair six-faced dice A and B be thrown simultaneously. If $E_{1}$ is the event that die A shows up four, $E_{2}$ is the event that die B shows up two, then which of the following statements is true?

Q 30 :

Two dice are rolled. If the probability of getting different numbers on these dice is $\frac{p}{q}$ where $p, q$ are relatively prime then $q - p$ is ____

Q 31 :

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$ , $g c d (m, n) = 1$ , then $m + n$ is equal to: [2026]

Q 32 :

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is: [2026]

Q 34 :

A bag contains $(N + 1)$ coins - $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$ , then $N$ is equal to : [2026]

Q 35 :

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is : [2026]

Q 37 :

Let $a, b, c \in {1, 2, 3, 4}$ . If the probability, that $a x^{2} + 2 \sqrt{2} b x + c > 0$ for all $x \in R$ , is $\frac{m}{n}$ , $\gcd (m, n) = 1$ , then $m + n$ is equal to ________. [2026]

Q 38 :

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $p$ , then $96 p$ is equal to ________. [2026]

Q 39 :

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$ , where $a, b \in N$ and $g c d (a, b) = 1$ , then $a + b$ is equal to ________. [2026]