Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, evaluate $P(A') + P(B')$.

Showing the 12 most recent of 32 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3$, then $P(A^c) + P(B^c) =$ (A) 0.3 (B) 0.5 (C) 0.7 (D) 0.9

   ›Reveal solutionSolution

   Use the complement rule and the inclusion–exclusion principle to express $P(A^c)+P(B^c)$ in terms of the given probabilities. The answer is 0.9.

   The key idea is that $P(A^c) = 1 - P(A)$ and $P(B^c) = 1 - P(B)$, so their sum is $2 - [P(A) + P(B)]$. We don’t know $P(A)$ or $P(B)$ individually, but we can find $P(A) + P(B)$ from the given $P(A \cup B)$ and $P(A \cap B)$ using the inclusion–exclusion formula.

   1. Recall the inclusion–exclusion principle For any two events $A$ and $B$,

   $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

   This is the fundamental relation that connects the union, intersection, and individual probabilities.

   2. Plug in the given values We have $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3$. Substituting:

   $0.8 = P(A) + P(B) - 0.3.$

   So

   $P(A) + P(B) = 0.8 + 0.3 = 1.1.$

   3. Express the required sum using complements The complement rule says $P(A^c) = 1 - P(A)$ and $P(B^c) = 1 - P(B)$. Therefore

   $P(A^c) + P(B^c) = [1 - P(A)] + [1 - P(B)] = 2 - [P(A) + P(B)].$

   4. Substitute the sum from step 2

   $P(A^c) + P(B^c) = 2 - 1.1 = 0.9.$

   Watch out

   A common mistake is to think $P(A^c) + P(B^c) = 1 - P(A \cup B)$ or something similar. That would be wrong — complements don’t combine that way. Always go back to the definition: $P(A^c) = 1 - P(A)$.

   Tip

   Notice that $P(A) + P(B)$ can exceed 1 (here it’s 1.1) because $A$ and $B$ overlap. That’s perfectly fine — it just means the events are not mutually exclusive.

   ✓Final answer

   The value is $\boxed{0.9}$, which corresponds to option (D).

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If A and B are any two events of a random experiment, then $\mathrm{P}\left[\left(A \cap B^c\right) \cup \left(A^c \cap B\right) \cup (A \cap B)\right]=$ (A) $\mathrm{P}(A) + \mathrm{P}(B)$ (B) $\mathrm{P}\left(A^c \cup B^c\right)$ (C) $1 - \mathrm{P}(A \cup B)$ (D) $\mathrm{P}(A \cup B)$

   ›Reveal solutionSolution

   The union of the three disjoint pieces — $A$ only, $B$ only, and both — is exactly the event that at least one of $A$ or $B$ occurs. So the probability is $\mathrm{P}(A \cup B)$, which is option (D).

   The question asks for the probability of a union of three set expressions. Before diving into algebra, notice what each piece represents:

   • $A \cap B^c$ : outcomes in $A$ but not in $B$ (only $A$).
   • $A^c \cap B$ : outcomes in $B$ but not in $A$ (only $B$).
   • $A \cap B$ : outcomes in both $A$ and $B$.

   These three sets are mutually disjoint — no outcome can belong to more than one of them at the same time. Their union therefore covers every outcome that belongs to $A$ or to $B$ (or to both). That is exactly the definition of $A \cup B$.

   So the whole expression simplifies immediately:

   $(A \cap B^c) \cup (A^c \cap B) \cup (A \cap B) = A \cup B.$

   Taking probability on both sides gives:

   $\mathrm{P}\left[(A \cap B^c) \cup (A^c \cap B) \cup (A \cap B)\right] = \mathrm{P}(A \cup B).$

   Now check the options:

   1. Option (A) $\mathrm{P}(A) + \mathrm{P}(B)$ is only correct when $A$ and $B$ are disjoint — not guaranteed here.
   2. Option (B) $\mathrm{P}(A^c \cup B^c)$ is the probability that at least one of them does not occur, which is $1 - \mathrm{P}(A \cap B)$ — not the same.
   3. Option (C) $1 - \mathrm{P}(A \cup B)$ is the probability that neither occurs — the complement of what we have.
   4. Option (D) $\mathrm{P}(A \cup B)$ matches exactly.

   Watch out

   A common mistake is to try expanding the union using inclusion-exclusion. That’s unnecessary here — the three pieces are already disjoint, so the probability of their union is simply the sum of their individual probabilities, which equals $\mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = \mathrm{P}(A \cup B)$.

   Tip

   Visualise a Venn diagram: the three regions — $A$ only, $B$ only, and the overlap — together form the entire $A \cup B$ region. No calculation needed.

   ✓Final answer

   The correct option is (D), $\mathrm{P}(A \cup B)$.

3. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If A and B are two events in a random experiment such that $P(A) + P(B) = 2P(A \cap B)$ then (A) $P(A) + P(B) = 1$ (B) $P(A) = P(B)$ (C) $P(A) + P(B) > 1$ (D) $P(A) = 0, P(B) = 1$

   ›Reveal solutionSolution

   The condition $P(A)+P(B)=2P(A\cap B)$ forces the two events to have equal probability, so the correct choice is (B).

   We start with the given equation:

   $P(A) + P(B) = 2P(A \cap B).$

   The key concept is the inclusion–exclusion principle for two events:

   $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

   This formula always holds, and probabilities lie between 0 and 1. The given condition is unusual because it ties the sum of the individual probabilities directly to the intersection. Our job is to see what restriction this places on $P(A)$ and $P(B)$.

   ---

   1. Rewrite the given condition using inclusion–exclusion. From $P(A) + P(B) = 2P(A \cap B)$, subtract $P(A \cap B)$ from both sides:

   $P(A) + P(B) - P(A \cap B) = P(A \cap B).$

   The left side is exactly $P(A \cup B)$, so we get:

   $P(A \cup B) = P(A \cap B).$

   2. Interpret what $P(A \cup B) = P(A \cap B)$ means.

      For any two events, $A \cap B \subseteq A \cup B$, so $P(A \cap B) \le P(A \cup B)$.

      Here they are equal, which implies that the set difference $(A \cup B) \setminus (A \cap B)$ has probability zero.

      In other words, the parts of $A$ and $B$ that are not in the overlap have zero probability.

      This forces $P(A \setminus B) = 0$ and $P(B \setminus A) = 0$.

   3. Conclude that $A$ and $B$ are essentially the same event (up to a null set).

      Since $P(A \setminus B) = 0$, we have $P(A) = P(A \cap B)$.

      Similarly, $P(B \setminus A) = 0$ gives $P(B) = P(A \cap B)$.

      Therefore:

   $P(A) = P(B) = P(A \cap B).$

   4. Check the options.
      • (A) $P(A)+P(B)=1$: Not forced; e.g., if $P(A)=P(B)=0.3$, then $P(A)+P(B)=0.6 \neq 1$.
      • (B) $P(A)=P(B)$: Yes, we just proved this.
      • (C) $P(A)+P(B)>1$: Not forced; could be less than 1.
      • (D) $P(A)=0, P(B)=1$: This would give $P(A)+P(B)=1$ and $P(A\cap B)=0$, so $1 = 2\cdot0$ is false.

   Thus the only necessary conclusion is that $P(A) = P(B)$.

   Tip

   A quick way to see it: From $P(A)+P(B)=2P(A\cap B)$, rearrange to $P(A)-P(A\cap B) = P(A\cap B)-P(B)$. The left side is $P(A\setminus B)$, the right side is $-(P(B\setminus A))$. Since both sides are nonnegative, they must be zero, giving $P(A)=P(A\cap B)=P(B)$.

   Watch out

   A common mistake is to assume $P(A\cap B)=0$ or $P(A)+P(B)=1$. Neither follows directly; the condition actually forces equality of the two probabilities, not a specific numeric value.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If A and B are two events of a random experiment such that $P(A \cup B) = P(A \cap B)$, then which one amongst the following four options is not true (A) A and B are equally likely (B) $P(A \cap B') = 0$ (C) $P(A' \cap B) = 0$ (D) $P(A) + P(B) = 1$

   ›Reveal solutionSolution

   The condition $P(A \cup B) = P(A \cap B)$ forces $A$ and $B$ to be identical events (up to probability zero), making them equally likely and their complements disjoint from each other — but it does not force their probabilities to sum to 1. Option (D) is the one that is not true.

   The key insight here is to translate the given equality into a relationship between the events themselves. For any two events, the union probability is always at least as large as the intersection probability — they are equal only when the part of the union that lies outside the intersection is empty (in a probabilistic sense).

   Let’s unpack that.

   1. Rewrite the condition using the addition rule. The standard formula is

   $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

   The problem gives $P(A \cup B) = P(A \cap B)$. Substitute:

   $P(A \cap B) = P(A) + P(B) - P(A \cap B).$

   Bring the $P(A \cap B)$ term from the right to the left:

   $2P(A \cap B) = P(A) + P(B).$

   So we have

   $P(A) + P(B) = 2P(A \cap B). \qquad(1)$

   2. Interpret what (1) means. Notice that $P(A) \ge P(A \cap B)$ and $P(B) \ge P(A \cap B)$. The only way their sum can be exactly twice the intersection is if each equals the intersection:

   $P(A) = P(A \cap B) \quad \text{and} \quad P(B) = P(A \cap B).$

   Why? Because if either $P(A) > P(A \cap B)$, then $P(A) + P(B) > 2P(A \cap B)$ (since $P(B) \ge P(A \cap B)$). The equality in (1) forces both to be exactly equal to the intersection.

   Hence

   $P(A) = P(B) = P(A \cap B).$

   3. Consequences of $P(A) = P(A \cap B)$. If $P(A) = P(A \cap B)$, then the part of $A$ that is not in $B$ has probability zero:

   $P(A \cap B') = P(A) - P(A \cap B) = 0.$

   Similarly, $P(B) = P(A \cap B)$ gives

   $P(A' \cap B) = P(B) - P(A \cap B) = 0.$

   So options (B) and (C) are true.

   4. Are A and B equally likely?

      From step 2, $P(A) = P(B)$. That is exactly what “equally likely” means in this context. So option (A) is true.

   5. What about option (D): $P(A) + P(B) = 1$?

      From (1), $P(A) + P(B) = 2P(A \cap B)$. There is no reason this must equal 1. For example, take a simple case: let $A = B$ be any event with probability $0.3$. Then $P(A \cup B) = 0.3$ and $P(A \cap B) = 0.3$, so the condition holds, but $P(A) + P(B) = 0.6 \neq 1$.

      So (D) is not forced by the given condition — it may be true in some special cases but is not generally true.

   Watch out

   A common mistake is to think $P(A \cup B) = P(A \cap B)$ implies $A = B$ as sets. That’s too strong — it only forces equality up to sets of probability zero. But for the purpose of checking the given options, the probabilistic equalities $P(A) = P(B)$ and the zero-probability complements are sufficient.

   ✓Final answer

   The option that is not true is (D).

5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A, B are the events in a random experiment. If $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$, $P(A \cap B) = \dfrac{1}{4}$, then $P\left(\dfrac{A^c}{B^c}\right) + P\left(\dfrac{A}{B}\right) =$ (A) $1$ (B) $\dfrac{4}{5}$ (C) $\dfrac{11}{8}$ (D) $\dfrac{7}{3}$

   ›Reveal solutionSolution

   The problem asks for the sum of two conditional probabilities: $P(A^c \mid B^c) + P(A \mid B)$. Using the definitions and given probabilities, we compute each term separately and add them. The result is $\frac{11}{8}$, which corresponds to option (C).

   We are given $P(A) = \frac12$, $P(B) = \frac13$, and $P(A \cap B) = \frac14$. The notation $P\left(\dfrac{A^c}{B^c}\right)$ means $P(A^c \mid B^c)$, the probability of $A$ not happening given that $B$ does not happen. Similarly, $P\left(\dfrac{A}{B}\right)$ is $P(A \mid B)$.

   The key idea: conditional probability is defined as $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$, provided $P(Y) > 0$. We will compute each conditional probability using the given data, then sum them.

   1. Compute $P(A \mid B)$ By definition:

   $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{1}{4} \cdot \frac{3}{1} = \frac{3}{4}.$

   2. Compute $P(A^c \mid B^c)$ First, find $P(B^c)$:

   $P(B^c) = 1 - P(B) = 1 - \frac13 = \frac23.$

   Next, find $P(A^c \cap B^c)$. By De Morgan’s law, $A^c \cap B^c = (A \cup B)^c$, so

   $P(A^c \cap B^c) = 1 - P(A \cup B).$

   We need $P(A \cup B)$:

   $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac12 + \frac13 - \frac14.$

   Get a common denominator of 12:

   $\frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12}.$

   Thus,

   $P(A^c \cap B^c) = 1 - \frac{7}{12} = \frac{5}{12}.$

   Now,

   $P(A^c \mid B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{5/12}{2/3} = \frac{5}{12} \cdot \frac{3}{2} = \frac{15}{24} = \frac{5}{8}.$

   3. Add the two conditional probabilities

   $P(A^c \mid B^c) + P(A \mid B) = \frac{5}{8} + \frac{3}{4} = \frac{5}{8} + \frac{6}{8} = \frac{11}{8}.$

   Tip

   A common mistake is to think $P(A^c \mid B^c) = 1 - P(A \mid B)$, but that is false in general. The complement rule for conditional probability is $P(A^c \mid B) = 1 - P(A \mid B)$, not when the condition changes.

   Watch out

   Be careful: $P(A^c \mid B^c)$ is not simply $1 - P(A \mid B^c)$; you must compute it directly from the definition.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A and B are two events of a random experiment such that P(B) = 0.4, P(A ∩ $\overline{B}$) = 0.5, P(A ∪ B) + P$\left(\dfrac{B}{A \cup \overline{B}}\right)$ = 1.15, then P(A) = (A) 0.9 (B) 0.8 (C) 0.7 (D) 0.25

   ›Reveal solutionSolution

   The key is to use the given probabilities and the conditional probability formula to set up an equation for $P(A)$. Solving yields $P(A) = 0.7$, so the correct option is (C).

   We are given:

   • $P(B) = 0.4$
   • $P(A \cap \overline{B}) = 0.5$
   • $P(A \cup B) + P\left(\dfrac{B}{A \cup \overline{B}}\right) = 1.15$

   We need $P(A)$.

   Concept and intuition:

   The problem mixes union, intersection, complement, and conditional probability. The key is to express everything in terms of $P(A)$ and known quantities. The conditional probability $P(B \mid A \cup \overline{B})$ can be rewritten using the definition:

   $P(B \mid A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$

   Then we simplify the numerator using set algebra. The given sum then becomes an equation in $P(A)$.

   Step-by-step solution:

   1. Express $P(A \cup B)$ in terms of $P(A)$. We know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Also, $P(A \cap \overline{B}) = P(A) - P(A \cap B)$. Given $P(A \cap \overline{B}) = 0.5$, we have

   $P(A) - P(A \cap B) = 0.5 \quad \Rightarrow \quad P(A \cap B) = P(A) - 0.5.$

   Therefore,

   $P(A \cup B) = P(A) + 0.4 - (P(A) - 0.5) = 0.9.$

   So $P(A \cup B) = 0.9$ — interestingly independent of $P(A)$! This is a key simplification.

   2. Find $P(A \cup \overline{B})$. Note that $A \cup \overline{B}$ is the complement of $B \cap \overline{A}$? Better: Use

   $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}).$

   We have $P(\overline{B}) = 1 - 0.4 = 0.6$ and $P(A \cap \overline{B}) = 0.5$. So

   $P(A \cup \overline{B}) = P(A) + 0.6 - 0.5 = P(A) + 0.1.$

   3. Compute the numerator for the conditional probability. We need $P(B \cap (A \cup \overline{B}))$. By distributive law:

   $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B}) = (A \cap B) \cup \varnothing = A \cap B.$

   So $P(B \cap (A \cup \overline{B})) = P(A \cap B) = P(A) - 0.5$ (from step 1).

   4. Write the conditional probability.

   $P\left(\frac{B}{A \cup \overline{B}}\right) = \frac{P(A \cap B)}{P(A \cup \overline{B})} = \frac{P(A) - 0.5}{P(A) + 0.1}.$

   5. Set up the given equation. We have

   $P(A \cup B) + \frac{P(A) - 0.5}{P(A) + 0.1} = 1.15.$

   Substitute $P(A \cup B) = 0.9$:

   $0.9 + \frac{P(A) - 0.5}{P(A) + 0.1} = 1.15.$

   Subtract 0.9:

   $\frac{P(A) - 0.5}{P(A) + 0.1} = 0.25.$

   6. Solve for $P(A)$. Multiply both sides by $P(A) + 0.1$:

   $P(A) - 0.5 = 0.25(P(A) + 0.1) = 0.25 P(A) + 0.025.$

   Bring terms:

   $P(A) - 0.25 P(A) = 0.5 + 0.025 \quad \Rightarrow \quad 0.75 P(A) = 0.525.$

   Hence

   $P(A) = \frac{0.525}{0.75} = 0.7.$

   Watch out

   A common mistake is to forget that $P(A \cup B)$ is not simply $P(A) + P(B)$ when events overlap. Here we used the given intersection with the complement to find the overlap correctly.

   Tip

   Notice that $P(A \cup B)$ turned out to be a constant (0.9) regardless of $P(A)$. This is because the given $P(A \cap \overline{B})$ fixed the relationship between $P(A)$ and $P(A \cap B)$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A and B are two events of a random experiment such that P(B) = 0.4, P(A ∩ $\overline{B}$) = 0.5, P(A ∪ B) + P$\left(\dfrac{B}{A \cup \overline{B}}\right)$ = 1.15, then P(A) = (A) 0.9 (B) 0.25 (C) 0.7 (D) 0.8

   ›Reveal solutionSolution

   The key is to use the given probability equation to solve for $P(A)$ by expressing everything in terms of $P(A)$ and $P(B)$, using set identities and conditional probability. The final value is $P(A) = 0.7$.

   The problem gives you a mix of basic probability and conditional probability. The trick is not to panic at the messy-looking term $P\left(\dfrac{B}{A \cup \overline{B}}\right)$ — that’s just conditional probability notation for $P(B \mid A \cup \overline{B})$. The equation $P(A \cup B) + P(B \mid A \cup \overline{B}) = 1.15$ is the main tool. You already know $P(B) = 0.4$ and $P(A \cap \overline{B}) = 0.5$. Your goal is to find $P(A)$.

   Let’s work through it step by step.

   1. Express $P(A \cup B)$ in terms of $P(A)$ and $P(B)$. The union formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. You don’t have $P(A \cap B)$ directly, but you have $P(A \cap \overline{B}) = 0.5$. Since $A$ is the disjoint union of $A \cap B$ and $A \cap \overline{B}$, we have:

   $P(A) = P(A \cap B) + P(A \cap \overline{B})$

   So $P(A \cap B) = P(A) - 0.5$.

   Therefore:

   $P(A \cup B) = P(A) + 0.4 - (P(A) - 0.5) = 0.9$

   Interesting — $P(A \cup B)$ simplifies to a constant $0.9$, independent of $P(A)$! That’s a neat simplification.

   2. Now handle the conditional probability term. $P(B \mid A \cup \overline{B})$ means the probability of $B$ happening, given that $A \cup \overline{B}$ has occurred. By definition:

   $P(B \mid A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$

   We need to simplify the numerator and denominator.

   3. Simplify $B \cap (A \cup \overline{B})$. Using distributive law: $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})$. But $B \cap \overline{B} = \emptyset$, so this is just $B \cap A = A \cap B$. Hence:

   $P(B \cap (A \cup \overline{B})) = P(A \cap B) = P(A) - 0.5$

   4. Simplify $P(A \cup \overline{B})$. Use the union formula: $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})$. $P(\overline{B}) = 1 - P(B) = 0.6$, and $P(A \cap \overline{B}) = 0.5$. So:

   $P(A \cup \overline{B}) = P(A) + 0.6 - 0.5 = P(A) + 0.1$

   5. Plug into the given equation. The equation is:

   $P(A \cup B) + P(B \mid A \cup \overline{B}) = 1.15$

   Substitute:

   $0.9 + \frac{P(A) - 0.5}{P(A) + 0.1} = 1.15$

   Subtract 0.9 from both sides:

   $\frac{P(A) - 0.5}{P(A) + 0.1} = 0.25$

   6. Solve for $P(A)$. Multiply both sides by $P(A) + 0.1$:

   $P(A) - 0.5 = 0.25(P(A) + 0.1)$

   $P(A) - 0.5 = 0.25P(A) + 0.025$

   Bring terms: $P(A) - 0.25P(A) = 0.5 + 0.025$

   $0.75P(A) = 0.525$

   $P(A) = \frac{0.525}{0.75} = 0.7$

   Watch out

   A common mistake is to forget that $P(A \cup \overline{B})$ is not simply $P(A) + P(\overline{B})$ — you must subtract the intersection $P(A \cap \overline{B})$, which is given. Also, note that $P(A \cup B)$ turned out to be constant; if you didn’t simplify it first, you might have gotten tangled.

   Tip

   The moment you saw $P(A \cup B)$ simplify to 0.9, you knew the conditional term had to be 0.25 to sum to 1.15. That shortcut saves time in an exam.

   ✓Final answer

   The value is $P(A) = 0.7$, which corresponds to option (C).

8. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.Two friends A and B meet every weekend either at a party or at a Sports Club. The probability that they meet at Sports Club is $\dfrac{4}{9}$. The probability that they will dine together at a party and at the Club are respectively $\dfrac{1}{3}$ and $\dfrac{2}{5}$. On a certain weekend the probability that they disperse without dine together (A) $\dfrac{86}{135}$ (B) $\dfrac{10}{27}$ (C) $\dfrac{17}{27}$ (D) $\dfrac{56}{135}$

   ›Reveal solutionSolution

   This problem asks for the total probability that two friends disperse without dining together, considering two possible meeting locations (party or sports club) and their respective conditional probabilities of dining. We use the concept of complementary events and the Law of Total Probability to sum the probabilities of not dining in each scenario. The final probability is $\boxed{\frac{86}{135}}$.

   The core idea in this problem is to use the Law of Total Probability. Since the friends must meet either at a party or at a sports club, these two events form a partition of the sample space. This means they are mutually exclusive (cannot happen at the same time) and exhaustive (cover all possibilities). We can calculate the probability of "not dining together" for each location separately and then sum these probabilities to get the overall probability.

   Here's how we break it down:

   1. Define Events and Given Probabilities:

      Let's clearly define the events involved to avoid confusion:

      • $S$: The friends meet at the Sports Club.
      • $P$: The friends meet at a Party.
      • $D_S$: The friends dine together at the Sports Club.
      • $D_P$: The friends dine together at a Party.

      From the problem statement, we are given the following probabilities:

      • The probability they meet at the Sports Club: $P(S) = \frac{4}{9}$.
      • Since they meet either at a party or at a Sports Club, these are the only two possibilities. Therefore, the probability they meet at a Party is the complement of meeting at the Sports Club: $P(P) = 1 - P(S) = 1 - \frac{4}{9} = \frac{5}{9}$.
      • The probability they dine together given they are at a party: $P(D_P | P) = \frac{1}{3}$.
      • The probability they dine together given they are at the Sports Club: $P(D_S | S) = \frac{2}{5}$.

   2. Calculate Probabilities of Not Dining Together (Conditional):

      We are interested in the event that they disperse without dining together. Let $D'$ denote this event.

      If they are at a party, the probability they do not dine together is the complement of dining together at the party:

      $P(D_P' | P) = 1 - P(D_P | P) = 1 - \frac{1}{3} = \frac{2}{3}$.

      Similarly, if they are at the Sports Club, the probability they do not dine together is the complement of dining together at the club:

      $P(D_S' | S) = 1 - P(D_S | S) = 1 - \frac{2}{5} = \frac{3}{5}$.

   3. Calculate Joint Probabilities of Not Dining Together:

      Now, we need to find the probability of two specific scenarios where they do not dine together:

      • Scenario 1: They meet at a Party and do not dine together there. This is the joint probability $P(P \cap D_P')$.
      • Scenario 2: They meet at the Sports Club and do not dine together there. This is the joint probability $P(S \cap D_S')$.

      We use the definition of conditional probability, which states $P(A \cap B) = P(B | A) \cdot P(A)$:

      For Scenario 1:

      $P(P \cap D_P') = P(D_P' | P) \cdot P(P) = \frac{2}{3} \cdot \frac{5}{9} = \frac{10}{27}$.

      For Scenario 2:

      $P(S \cap D_S') = P(D_S' | S) \cdot P(S) = \frac{3}{5} \cdot \frac{4}{9} = \frac{12}{45}$.

      This fraction can be simplified by dividing both the numerator and denominator by their greatest common divisor, 3: $\frac{12 \div 3}{45 \div 3} = \frac{4}{15}$.

   4. Apply the Law of Total Probability:

      The event "they disperse without dining together" ($D'$) can occur in one of two mutually exclusive ways: either they meet at a Party and don't dine, or they meet at the Sports Club and don't dine. Since these are the only two possibilities for where they meet, we can sum their probabilities to find the total probability of $D'$.

      The Law of Total Probability states that if $B_1, B_2, \dots, B_n$ are mutually exclusive and exhaustive events, then for any event $A$:

      $P(A) = \sum_{i=1}^n P(A | B_i) P(B_i)$

      In our case, $A$ is $D'$ (not dining together), and $B_1$ is $P$ (meeting at a party), $B_2$ is $S$ (meeting at a sports club).

      So, $P(D') = P(D' | P) P(P) + P(D' | S) P(S)$.

      Note that $P(D' | P)$ is $P(D_P' | P)$ and $P(D' | S)$ is $P(D_S' | S)$.

      This is equivalent to $P(D') = P(P \cap D_P') + P(S \cap D_S')$.

      Using the calculated joint probabilities:

      $P(D') = \frac{10}{27} + \frac{4}{15}$.

      To add these fractions, we find the least common multiple (LCM) of the denominators 27 and 15.

      $27 = 3^3$

      $15 = 3 \times 5$

      The $LCM(27, 15) = 3^3 \times 5 = 27 \times 5 = 135$.

      Now, we convert the fractions to have this common denominator:

      $\frac{10}{27} = \frac{10 \times 5}{27 \times 5} = \frac{50}{135}$

      $\frac{4}{15} = \frac{4 \times 9}{15 \times 9} = \frac{36}{135}$

      Finally, we add the fractions:

      $P(D') = \frac{50}{135} + \frac{36}{135} = \frac{50 + 36}{135} = \frac{86}{135}$.

   ✓Final answer

   The probability that the friends disperse without dining together is $\boxed{\frac{86}{135}}$.

9. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If A, B, C are three mutually exclusive and exhaustive events such that P(A):P(B):P(C) = 1:l:m, then P(A \cup B) + P(B \cup C) + P(C \cup A) + P(A \cup B \cup C) = (A) $\dfrac{1}{3}$ (B) $3$ (C) $\dfrac{3}{4}$ (D) $1$

   ›Reveal solutionSolution

   The key idea is to express all probabilities in terms of a single unknown using the given ratio, then apply the inclusion-exclusion principle for three mutually exclusive and exhaustive events. The sum simplifies to a constant independent of the ratio, giving the answer 3.

   We are told that A, B, C are mutually exclusive (no two can happen at once) and exhaustive (together they cover the whole sample space). That means:

   • $P(A \cap B) = P(B \cap C) = P(C \cap A) = 0$
   • $P(A \cup B \cup C) = 1$

   The ratio $P(A):P(B):P(C) = 1 : l : m$ is given, but note that $l$ and $m$ are just positive numbers (not necessarily integers). Since the events are exhaustive, the sum of their probabilities is 1.

   Let’s work through the problem step by step.

   1. Set up the probabilities using the ratio. Let $P(A) = k$. Then from the ratio, $P(B) = lk$ and $P(C) = mk$. Because the events are exhaustive:

   $P(A) + P(B) + P(C) = k + lk + mk = k(1 + l + m) = 1$

   So:

   $k = \frac{1}{1 + l + m}$

   2. Interpret the required expression. We need:

   $S = P(A \cup B) + P(B \cup C) + P(C \cup A) + P(A \cup B \cup C)$

   Since A, B, C are mutually exclusive, the union of any two is just the sum of their probabilities. For example:

   $P(A \cup B) = P(A) + P(B) \quad \text{(no overlap)}$

   Similarly for the other pairs. And $P(A \cup B \cup C) = 1$ because they are exhaustive.

   3. Substitute these simplifications.

   $S = [P(A) + P(B)] + [P(B) + P(C)] + [P(C) + P(A)] + 1$

   Count the occurrences: $P(A)$ appears twice, $P(B)$ appears twice, $P(C)$ appears twice. So:

   $S = 2[P(A) + P(B) + P(C)] + 1$

   4. Use the exhaustive property. Since $P(A) + P(B) + P(C) = 1$, we get:

   $S = 2 \times 1 + 1 = 3$

   Tip

   The ratio $1:l:m$ is a red herring — the sum simplifies to a constant regardless of the specific values of $l$ and $m$. This is because the exhaustive property forces the total probability to 1, and mutual exclusivity makes the unions additive.

   Watch out

   A common mistake is to try to compute each union using inclusion-exclusion with intersections, but since the events are mutually exclusive, all intersections are empty. The expression collapses to a simple sum.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.If 2 coins are tossed and 2 dice are thrown at a time, then the probability of getting atleast 1 head and the sum of the numbers appeared on the dice as atleast 9 is (A) $\dfrac{5}{36}$ (B) $\dfrac{1}{6}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{5}{24}$

    ›Reveal solutionSolution

    The probability is the product of the independent coin and dice events: P(at least 1 head) = 3/4, P(sum ≥ 9) = 5/18, so the combined probability is (3/4)×(5/18) = 5/24, which corresponds to option (D).

    We have two independent experiments: tossing two coins and throwing two dice. Because they are independent, the probability of both events happening is simply the product of their individual probabilities. The key is to compute each separately, then multiply.

    1. Probability of at least 1 head from two coins

       • Total outcomes when tossing two coins: $2^2 = 4$ (HH, HT, TH, TT).
       • "At least 1 head" means we exclude the case of no heads (TT).
       • Number of favorable outcomes = 3.
       • So $P(\text{at least 1 head}) = \frac{3}{4}$.

    2. Probability that the sum of two dice is at least 9

       • Total outcomes when throwing two dice: $6 \times 6 = 36$.
       • Sums that are at least 9: 9, 10, 11, 12.
       • Count the number of ways for each sum:
         • Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 ways.
         • Sum = 10: (4,6), (5,5), (6,4) → 3 ways.
         • Sum = 11: (5,6), (6,5) → 2 ways.
         • Sum = 12: (6,6) → 1 way.
       • Total favorable = $4 + 3 + 2 + 1 = 10$.
       • So $P(\text{sum} \geq 9) = \frac{10}{36} = \frac{5}{18}$.

    3. Combine the independent probabilities

       • Since the coin toss and dice throw are independent, multiply:

    $P = \frac{3}{4} \times \frac{5}{18} = \frac{15}{72} = \frac{5}{24}.$

    Watch out

    A common mistake is to forget that the two events are independent and try to list all combined outcomes (2 coins × 2 dice = 4 × 36 = 144 outcomes) and count manually — which works but is slower. The product rule is much faster.

    Tip

    Notice that "at least 1 head" is the complement of "no heads", which is $1 - \frac{1}{4} = \frac{3}{4}$. Similarly, "sum at least 9" can be found by symmetry or by listing the 10 favorable pairs out of 36.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.A bag contains 3 red, 5 black and 7 blue balls. If three balls are drawn at random simultaneously from the bag then the probability of getting at least two blue balls is (A) $\dfrac{29}{65}$ (B) $\dfrac{29}{130}$ (C) $\dfrac{9}{65}$ (D) $\dfrac{9}{130}$

    ›Reveal solutionSolution

    The probability of drawing at least two blue balls from a bag of 3 red, 5 black, and 7 blue balls (total 15) when three are drawn simultaneously is found by summing the cases of exactly two blues and exactly three blues. The result is $\frac{29}{65}$, which corresponds to option (A).

    We want the probability that among three balls drawn at random from the bag, at least two are blue. "At least two" means either exactly two blue balls or all three blue balls. Since the draws are simultaneous, we use combinations (order doesn't matter). The total number of ways to choose any three balls from the 15 is $\binom{15}{3}$. The favorable cases are counted by choosing the required number of blue balls from the 7 blue, and the rest from the non-blue balls (3 red + 5 black = 8 non-blue).

    1. Total number of outcomes Total balls = $3 + 5 + 7 = 15$. Number of ways to choose any 3 balls:

    $\binom{15}{3} = \frac{15 \cdot 14 \cdot 13}{3 \cdot 2 \cdot 1} = 455.$

    2. Case 1: Exactly two blue balls Choose 2 blue from the 7 blue: $\binom{7}{2} = 21$ ways. Choose the remaining 1 ball from the 8 non-blue: $\binom{8}{1} = 8$ ways. So number of favorable outcomes for exactly two blues:

    $21 \times 8 = 168.$

    3. Case 2: Exactly three blue balls Choose 3 blue from the 7 blue: $\binom{7}{3} = 35$ ways. No non-blue balls needed. So number of favorable outcomes for three blues:

    $35.$

    4. Total favorable outcomes

    $168 + 35 = 203.$

    5. Probability

    $P(\text{at least two blue}) = \frac{203}{455}.$

    Simplify the fraction: divide numerator and denominator by 7:

    $\frac{203 \div 7}{455 \div 7} = \frac{29}{65}.$

    Watch out

    A common mistake is to forget that "at least two" includes the case of three blues, or to treat the draws as ordered (permutations) and then forget to adjust. Using combinations avoids this pitfall entirely.

    Tip

    Notice that $203/455$ simplifies neatly because 203 = 7 × 29 and 455 = 7 × 65. Always check for common factors before finalizing.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.A and B are the two groups of books. Group A consists of 8 science and 5 engineering books and the group B consists of 6 science and 7 engineering books. When an unbiased die is rolled, if 2 or 5 turns up, a book is selected at random from the group A, otherwise a book is selected at random from the B group. When an unbiased die is rolled, the probability of selecting a science book is (A) $\dfrac{13}{24}$ (B) $\dfrac{34}{35}$ (C) $\dfrac{20}{39}$ (D) $\dfrac{13}{36}$

    ›Reveal solutionSolution

    Use the law of total probability: the overall chance of picking a science book is the weighted average of the science-book probabilities from each group, with weights given by the die-roll outcomes. The answer is $\frac{20}{39}$.

    The core idea here is that the selection happens in two stages: first the die decides which group you draw from, then you pick a book at random from that group. When a problem has this "choose a source, then pick from it" structure, the law of total probability is your natural tool. You break the overall event (picking a science book) into the mutually exclusive ways it can happen — via group A or via group B — and add their probabilities.

    Let’s walk through it.

    1. Determine the probabilities from the die roll. An unbiased die has six faces: 1, 2, 3, 4, 5, 6.
       • If 2 or 5 turns up, we select from group A. That’s 2 favourable outcomes out of 6.

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

    • For any other outcome (1, 3, 4, 6), we select from group B. That’s 4 outcomes.

    $P(B) = \frac{4}{6} = \frac{2}{3}$

    2. Find the probability of picking a science book from each group.
       • Group A has 8 science and 5 engineering books, so 13 books total.

    $P(\text{science} \mid A) = \frac{8}{13}$

    • Group B has 6 science and 7 engineering books, so 13 books total.

    $P(\text{science} \mid B) = \frac{6}{13}$

    Notice both groups have the same total number of books — that’s a coincidence, not a rule.

    3. Apply the law of total probability. The overall probability of selecting a science book is:

    $P(\text{science}) = P(A) \cdot P(\text{science} \mid A) + P(B) \cdot P(\text{science} \mid B)$

    Substitute the values:

    $P(\text{science}) = \frac{1}{3} \cdot \frac{8}{13} + \frac{2}{3} \cdot \frac{6}{13}$

    $= \frac{8}{39} + \frac{12}{39}$

    $= \frac{20}{39}$

    Watch out

    A common mistake is to forget that the die probabilities are not equal — many students assume each group is equally likely because the problem doesn’t explicitly say "biased". Always count the die faces carefully: 2 and 5 are only two faces out of six, so group A gets a probability of $1/3$, not $1/2$.

    Tip

    When both groups have the same total number of books (here, 13 each), you can think of the overall probability as a weighted average of the fractions $\frac{8}{13}$ and $\frac{6}{13}$ with weights $\frac{1}{3}$ and $\frac{2}{3}$. That’s exactly what we did.

    ✓Final answer

    The probability of selecting a science book is $\boxed{\frac{20}{39}}$, which corresponds to option (C).