If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P(A B) = 0.5, where B denotes the complement of B, then…

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Questions	9	17	9	35

Q jee_main_2024_30_january_evening Bayes Theorem

Bag A contains

3

white,

7

red balls and bag B contains

3

white,

2

red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is :

A. $(1)/(4)$ $\frac{1}{4}$
B. $(1)/(9)$ $\frac{1}{9}$
C. $(1)/(3)$ $\frac{1}{3}$
D. $(3)/(10)$ $\frac{3}{10}$

Solution

Related Formula

Bayes' Theorem: P(E₁ | E) = (P(E₁)P(E | E₁))/(P(E₁)P(E | E₁) + P(E₂)P(E | E₂))

Core Logic

Let $E₁$ $E_1$ be the event that Bag A is selected, and $E₂$ $E_2$ be the event that Bag B is selected.

P(E₁) = P(E₂) = (1)/(2)

Let $E$ $E$ be the event that a white ball is drawn. From Bag A (3 white, 7 red, total 10): $P(E | E₁) = (3)/(10)$ $P(E | E_1) = \frac{3}{10}$ From Bag B (3 white, 2 red, total 5): $P(E | E₂) = (3)/(5)$ $P(E | E_2) = \frac{3}{5}$

Step 1: Calculating the Target Probability

We need to find the probability that the ball was drawn from Bag A given it is white, i.e., $P(E₁ | E)$ $P(E_1 | E)$.

P(E₁ | E) = ((1)/(2) \times (3)/(10))/((1)/(2) \times (3)/(10) + (1)/(2) \times (3)/(5))

Canceling out $(1)/(2)$ $\frac{1}{2}$ from the numerator and the denominator:

P(E₁ | E) = ((3)/(10))/((3)/(10) + (6)/(10)) = (3)/(3 + 6) = (3)/(9) = (1)/(3)

Pattern Recognition

Reverse probability with disjoint prior states directly signals Bayes' theorem. Canceling prior probability terms ( $P(E₁)=P(E₂)$ $P(E_1)=P(E_2)$) speeds up the calculation.

Chapter Mix

Class 12 Maths: Probability

Q11 jee_main_2024_30_jan_morning Classical Probability

Two integers

x

and

y

are chosen with replacement from the set

0, 1, 2, 3, , 10

. Then the probability that

|x - y| > 5

is:

A. $(30)/(121)$ $\frac{30}{121}$
B. $(62)/(121)$ $\frac{62}{121}$
C. $(60)/(121)$ $\frac{60}{121}$
D. $(31)/(121)$ $\frac{31}{121}$

Solution

Related Formula

P(E) = Number of favorable outcomesTotal number of possible outcomes

Core Logic

Total possible selections for $(x, y)$ $(x, y)$ with replacement from $0, 1, , 10$ ${0, 1, \dots, 10}$ is $11 \times 11 = 121$ $11 \times 11 = 121$. We need pairs $(x, y)$ $(x, y)$ such that $|x - y| > 5$ $|x - y| > 5$, which means $x - y > 5$ $x - y > 5$ or $y - x > 5$ $y - x > 5$.

Step 1: Counting Favorable Cases

Assume $x < y$ $x < y$, so we need $y - x \geq 6$ $y - x \geq 6$. If $x = 0 \Rightarrow y in 6, 7, 8, 9, 10$ (5 ways) If $x = 1 \Rightarrow y in 7, 8, 9, 10$ (4 ways) If $x = 2 \Rightarrow y in 8, 9, 10$ $x = 2 \Rightarrow y \in \{8, 9, 10\}$ (3 ways) If $x = 3 \Rightarrow y in 9, 10$ $x = 3 \Rightarrow y \in \{9, 10\}$ (2 ways) If $x = 4 \Rightarrow y in 10$ $x = 4 \Rightarrow y \in \{10\}$ (1 way) If $x \geq 5$ $x \ge 5$, there are no possible values for $y$ $y$ strictly greater than $x$ $x$ satisfying the condition.

Step 2: Total Probability

The number of cases for $y > x$ $y > x$ is $5 + 4 + 3 + 2 + 1 = 15$ $5 + 4 + 3 + 2 + 1 = 15$. By symmetry, the number of cases for $x > y$ $x > y$ is also $15$ $15$. Total favorable cases $= 15 \times 2 = 30$ $= 15 \times 2 = 30$. Required probability $= (30)/(121)$ $= \frac{30}{121}$.

Pattern Recognition

Absolute difference conditions $|x - y| > k$ $|x - y| > k$ on discrete sets cleanly split into symmetric additive series $1+2+...+n$ $1+2+...+n$. Calculate one half and multiply by 2.

Chapter Mix

Class 12 Maths: Probability

Q18 jee_main_2024_31_jan_evening Binomial Distribution / Independent Events

A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

A. $(2)/(9)$ $\frac{2}{9}$
B. $(1)/(9)$ $\frac{1}{9}$
C. $(2)/(27)$ $\frac{2}{27}$
D. $(1)/(27)$ $\frac{1}{27}$

Solution

Related Formula

P(X=k) = ⁿCk \cdot p^k \cdot qn-k

Core Logic

Given $P(H) = 2P(T)$ $P(H) = 2P(T)$. Since $P(H) + P(T) = 1$ $P(H) + P(T) = 1$, we get $2P(T) + P(T) = 1 3P(T) = 1 P(T) = (1)/(3)$ . Then, $P(H) = (2)/(3)$ $P(H) = \frac{2}{3}$.

The coin is tossed 3 times. We need the probability of getting exactly 2 tails and 1 head. Using binomial probability:

P(2T, 1H) = ³C₂ \times (P(T))² \times (P(H))¹

= 3 \times ((1)/(3))² \times ((2)/(3))

= 3 \times (1)/(9) \times (2)/(3) = (6)/(27) = (2)/(9)

Chapter Mix

Class 12 Maths: Probability

Q16 jee_main_2024_31_jan_morning Independent Events

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

A. $(2)/(25)$ $$\frac{2}{25}$$
B. $(4)/(25)$ $$\frac{4}{25}$$
C. $(2)/(3)$ $$\frac{2}{3}$$
D. $(4)/(75)$ $$\frac{4}{75}$$

Solution

Core Logic

Total marbles $= 10 + 30 + 20 + 15 = 75$ $= 10 + 30 + 20 + 15 = 75$. Drawings are made with replacement, so the events are independent.

Step 1: Probability Calculation

Probability of first drawing a red marble: $P(R) = (10)/(75)$ $P(R) = \frac{10}{75}$. Probability of second drawing a white marble: $P(W) = (30)/(75)$ $P(W) = \frac{30}{75}$. Since they are independent: $P(R and W) = (10)/(75) \times (30)/(75) = (4)/(75)$ .

Chapter Mix

Class 12 Maths: Probability

Q19 jee_main_2024_31_jan_morning Variance of Random Variable

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable

X

to be the number of rotten apples in a draw of two apples, the variance of

X

is

A. $(37)/(153)$ $$\frac{37}{153}$$
B. $(57)/(153)$ $$\frac{57}{153}$$
C. $(47)/(153)$ $$\frac{47}{153}$$
D. $(40)/(153)$ $$\frac{40}{153}$$

Solution

Core Logic

Total apples = 18 (3 rotten, 15 good). Random variable $X = 0, 1, 2$ $X = \{0, 1, 2\}$ representing the number of rotten apples.

Step 1: Probability Distribution

P(X = 0) = ¹⁵C₂¹⁸C₂ = (105)/(153)

P(X = 1) = ³C₁ \times ¹⁵C₁¹⁸C₂ = (45)/(153)

P(X = 2) = ³C₂¹⁸C₂ = (3)/(153)

Step 2: Expectation

E(X) = 0 \times (105)/(153) + 1 \times (45)/(153) + 2 \times (3)/(153) = (51)/(153) = (1)/(3)

Step 3: Variance

E(X²) = 0 \times (105)/(153) + 1 \times (45)/(153) + 4 \times (3)/(153) = (57)/(153)

Var(X) = E(X²) - (E(X))² = (57)/(153) - ((1)/(3))² = (57)/(153) - (17)/(153) = (40)/(153)

Chapter Mix

Class 12 Maths: Probability

If A and B are two events such that P(A) = 0.7$P(A) = 0.7$, P(B) = 0.4$P(B) = 0.4$ and P(A B) = 0.5$P(A \cap \overline{B}) = 0.5$, where B$\overline{B}$ denotes the complement of B, then P(B | (A B))$P(B | (A \cup \overline{B}))$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Component Intersections

Step 2: Calculate Set Union

Step 3: Resolve Final Conditional Probability

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Calculating the Target Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Counting Favorable Cases

Step 2: Total Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Probability Calculation

Chapter Mix

Solution

Core Logic

Step 1: Probability Distribution

Step 2: Expectation

Step 3: Variance

Chapter Mix

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If A and B are two events such that $P(A) = 0.7$ $P(A) = 0.7$, $P(B) = 0.4$ $P(B) = 0.4$ and $P(A B) = 0.5$ $P(A \cap \overline{B}) = 0.5$, where $B$ $\overline{B}$ denotes the complement of B, then $P(B | (A B))$ $P(B | (A \cup \overline{B}))$ is equal to: