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:

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