Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :

Related Formula

Probability of an event P(E') = 1 - P(E)$P(E') = 1 - P(E)$, where E$E$ is the complementary event.

Core Logic

The word GARDEN contains 6 distinct letters: G, A, R, D, E, N. Total number of permutations (words in set S) = 6! = 720$6! = 720$.

The vowels present are A and E. In any random arrangement of these letters, there are only 2 possible mutual relative arrangements for the vowels:

• A appears before E (alphabetical order)
• E appears before A

By symmetry, both relative arrangements are equally probable.

Step 1: Calculate Probabilities

Probability that vowels are in alphabetical order = (1)/(2)$\frac{1}{2}$.

Therefore, the probability that the selected word will NOT have vowels in alphabetical order is:

Pattern Recognition

Symmetry Shortcut: For any k$k$ distinct specific objects inside an arrangement of distinct items, the number of ways they can be sorted in a unique relative order is exactly 1/k!$1/k!$ of the total arrangements. Here k=2$k=2$ vowels, so probability of alphabetical order is 1/2! = 1/2$1/2! = 1/2$. Not alphabetical is 1 - 1/2 = 1/2$1 - 1/2 = 1/2$.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability

Related Formula

According to Bayes' Theorem, the conditional probability of an event Ek$E_k$ given that event A$A$ has occurred is:

Core Logic

Let A$A$ be the event that 2$2$ white and 2$2$ black balls are drawn from the bag containing 8$8$ balls.

Since 2$2$ white and 2$2$ black balls have already been drawn, the initial composition of the bag could only be one of the following configurations:

• E₁$E_1$: 2$2$ White, 6$6$ Black
• E₂$E_2$: 3$3$ White, 5$5$ Black
• E₃$E_3$: 4$4$ White, 4$4$ Black (Equal composition)
• E₄$E_4$: 5$5$ White, 3$3$ Black
• E₅$E_5$: 6$6$ White, 2$2$ Black

Assuming these 5$5$ configurations are equally likely initially:

Step 1: Compute Conditional Probabilities

We calculate the probability of drawing 2$2$ white and 2$2$ black balls under each hypothesis using combinations:

• For E₁$E_1$ (2W, 6B$2\text{W}, 6\text{B}$): P(A|E₁) = ²C₂ · ⁶C₂⁸C₄ = (1 · 15)/(70) = (15)/(70)$P(A|E_1) = \frac{^{2}C_{2} \cdot ^{6}C_{2}}{^{8}C_{4}} = \frac{1 \cdot 15}{70} = \frac{15}{70}$
• For E₂$E_2$ (3W, 5B$3\text{W}, 5\text{B}$): P(A|E₂) = ³C₂ · ⁵C₂⁸C₄ = (3 · 10)/(70) = (30)/(70)$P(A|E_2) = \frac{^{3}C_{2} \cdot ^{5}C_{2}}{^{8}C_{4}} = \frac{3 \cdot 10}{70} = \frac{30}{70}$
• For E₃$E_3$ (4W, 4B$4\text{W}, 4\text{B}$): P(A|E₃) = ⁴C₂ · ⁴C₂⁸C₄ = (6 · 6)/(70) = (36)/(70)$P(A|E_3) = \frac{^{4}C_{2} \cdot ^{4}C_{2}}{^{8}C_{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70}$
• For E₄$E_4$ (5W, 3B$5\text{W}, 3\text{B}$): P(A|E₄) = ⁵C₂ · ³C₂⁸C₄ = (10 · 3)/(70) = (30)/(70)$P(A|E_4) = \frac{^{5}C_{2} \cdot ^{3}C_{2}}{^{8}C_{4}} = \frac{10 \cdot 3}{70} = \frac{30}{70}$
• For E₅$E_5$ (6W, 2B$6\text{W}, 2\text{B}$): P(A|E₅) = ⁶C₂ · ²C₂⁸C₄ = (15 · 1)/(70) = (15)/(70)$P(A|E_5) = \frac{^{6}C_{2} \cdot ^{2}C_{2}}{^{8}C_{4}} = \frac{15 \cdot 1}{70} = \frac{15}{70}$

Step 2: Apply Bayes' Theorem

We want to find P(E₃|A)$P(E_3|A)$, the probability that the bag contains equal numbers of white and black balls:

Pattern Recognition

Sees: Total number of balls is known, and a sample outcome is given to find the initial state distribution. Shortcut: Notice the symmetry in the configuration possibilities (E₁$E_1$ and E₅$E_5$ have identical probabilities, as do E₂$E_2$ and E₄$E_4$). Sum the numerator terms directly without writing out the full expansion to save crucial seconds.

Chapter Mix

Class 12 Mathematics: Probability

Related Formula

Core Logic

Let total sample space S = 1, 2, , 50 n(S) = 50$S = \{1, 2, \dots, 50\} \implies n(S) = 50$.

• Let A$A$ be multiple of 4: 4, 8, , 48 n(A) = (50)/(4) = 12$4, 8, \dots, 48 \implies n(A) = \lfloor \frac{50}{4} \rfloor = 12$
• Let B$B$ be multiple of 6: 6, 12, , 48 n(B) = (50)/(6) = 8$6, 12, \dots, 48 \implies n(B) = \lfloor \frac{50}{6} \rfloor = 8$
• Let C$C$ be multiple of 7: 7, 14, , 49 n(C) = (50)/(7) = 7$7, 14, \dots, 49 \implies n(C) = \lfloor \frac{50}{7} \rfloor = 7$

Step 1: Finding Intersections

• A B$A \cap B$ (LCM of 4 and 6 = 12): multiples of 12 n(A B) = (50)/(12) = 4$\implies n(A \cap B) = \lfloor \frac{50}{12} \rfloor = 4$
• B C$B \cap C$ (LCM of 6 and 7 = 42): multiples of 42 n(B C) = (50)/(42) = 1$\implies n(B \cap C) = \lfloor \frac{50}{42} \rfloor = 1$
• A C$A \cap C$ (LCM of 4 and 7 = 28): multiples of 28 n(A C) = (50)/(28) = 1$\implies n(A \cap C) = \lfloor \frac{50}{28} \rfloor = 1$
• A B C$A \cap B \cap C$ (LCM of 4, 6, 7 = 84): multiples of 84 n(A B C) = 0$\implies n(A \cap B \cap C) = 0$

Step 2: Final Enumeration

Applying the Principle of Inclusion-Exclusion:

Therefore, the probability is:

Pattern Recognition

To quickly count the number of multiples up to N$N$, use the greatest integer function NLCM$\lfloor \frac{N}{\text{LCM}} \rfloor$. This systematically prevents manual counting blunders.

Chapter Mix

Class 11 Mathematics: Probability

Related Formula

Core Logic

This is a geometric probability distribution. Probability of success (rolling a 6) is p = (1)/(6)$p = \frac{1}{6}$, and failure is q = (5)/(6)$q = \frac{5}{6}$.

Calculate a = P(X=3)$a = P(X=3)$: This means the first two tosses are failures, and the third is a success.

Step 1: Evaluating Cumulative Probability b

Calculate b = P(X ≥ 3)$b = P(X \ge 3)$: This means the first two tosses must be failures (what happens after doesn't matter).

Step 2: Evaluating Conditional Probability c

Calculate c = P(X ≥ 6 | X > 3)$c = P(X \ge 6 | X > 3)$: By conditional probability formula:

Using the logic from b$b$:

Step 3: Final Computation

Now compute the requested value (b+c)/(a)$\frac{b+c}{a}$:

Pattern Recognition

The geometric distribution is "memoryless." Hence, the conditional probability P(X ≥ k+m | X > m)$P(X \ge k+m | X > m)$ is identical to the unconditional probability P(X ≥ k)$P(X \ge k)$. Here, P(X ≥ 6 | X ≥ 4)$P(X \ge 6 | X \ge 4)$ directly equals P(X ≥ 3) = b$P(X \ge 3) = b$.

Chapter Mix

Class 12 Maths: Probability

Related Formula

Where S∞$S_{\infty}$ is the sum of an infinite geometric progression, a$a$ is the first term, and r$r$ is the common ratio.

Core Logic

Let success (S$S$) be rolling a 2, and failure (F$F$) be rolling anything else. P(S) = (1)/(6)$P(S) = \frac{1}{6}$ P(F) = (5)/(6)$P(F) = \frac{5}{6}$

We want the probability that the first success occurs on an even number of throws (2nd, 4th, 6th, $\dots$). The sequence of events for success on even throws is:

• Success on 2nd throw: F, S$F, S$
• Success on 4th throw: F, F, F, S$F, F, F, S$
• Success on 6th throw: F, F, F, F, F, S$F, F, F, F, F, S$

Writing this as a sum of probabilities:

Step 1: Compute Infinite Series Sum

This is an infinite geometric series with: First term a = (5)/(6) × (1)/(6) = (5)/(36)$a = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ Common ratio r = ((5)/(6))² = (25)/(36)$r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

Applying the sum formula:

Pattern Recognition

For alternating success/failure probabilities P(Even) = (q · p)/(1 - q²)$P(\text{Even}) = \frac{q \cdot p}{1 - q^2}$ and P(Odd) = (p)/(1 - q²)$P(\text{Odd}) = \frac{p}{1 - q^2}$. Knowing this format immediately turns it into a 5-second mental calculation: ((5/6)(1/6))/(1 - 25/36) = 5/11$\frac{(5/6)(1/6)}{1 - 25/36} = 5/11$.

Chapter Mix

Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series