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)

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

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

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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)

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

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

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

$Similarly,$

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

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

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

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

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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

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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