Q.State whether the following statement is True or False: If $A$ and $B$ are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$, then $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$.

Showing the 12 most recent of 32 on this concept.

1. 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. …

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2. 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? …

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3. 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, …

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4. 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 …

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5. 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: …

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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 $$$$

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7. 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. …

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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: …

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

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10. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.A bag A contains 3 red, 2 white and 2 black balls and another bag B contains 1 red, 2 white and 4 black balls. A die is thrown to select a bag from which a ball has to be chosen. If an odd prime number appears on the die, a ball is drawn from bag A; otherwise, a ball is drawn from bag B. With this condition, if the ball drawn is found to be black, then the probability that it is drawn from bag B is (A) $\dfrac{6}{7}$ (B) $\dfrac{3}{7}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{4}{5}$

    ›Reveal solutionSolution

    Use Bayes' theorem to reverse the conditional probability: the chance that the black ball came from bag B is $\frac{4}{5}$.

    The problem is a classic Bayes' theorem setup. We are told the outcome (a black ball) and need the probability that it came from a particular source (bag B). The die decides which bag is chosen first, so we have prior probabilities for each bag. Then, given the bag, we know the chance of drawing a black ball. Bayes' theorem lets us flip the condition: from "probability of black given bag" to "probability of bag given black."

    Let’s define the events clearly:

    • $A$: bag A is chosen
    • $B$: bag B is chosen
    • $Bl$: a black ball is drawn

    The die: an odd prime number on a die is 3 or 5 (since 2 is prime but even, and 1 is not prime). So odd primes are 3 and 5 — that's 2 outcomes out of 6.

    1. Prior probabilities

       $P(A) = \frac{2}{6} = \frac{1}{3}$ (when die shows 3 or 5)

       $P(B) = 1 - \frac{1}{3} = \frac{2}{3}$ (when die shows 1, 2, 4, or 6)

    2. Likelihoods — probability of drawing a black ball from each bag

       Bag A: 3 red, 2 white, 2 black → total 7 balls, so $P(Bl \mid A) = \frac{2}{7}$

       Bag B: 1 red, 2 white, 4 black → total 7 balls, so $P(Bl \mid B) = \frac{4}{7}$

    3. Apply Bayes' theorem

       We want $P(B \mid Bl)$, the probability that the ball came from bag B given it is black.

       Bayes' theorem says:

    $P(B \mid Bl) = \frac{P(Bl \mid B) \cdot P(B)}{P(Bl)}$

    The denominator $P(Bl)$ is the total probability of drawing a black ball:

    $P(Bl) = P(Bl \mid A)P(A) + P(Bl \mid B)P(B)$

    $= \left(\frac{2}{7}\right)\left(\frac{1}{3}\right) + \left(\frac{4}{7}\right)\left(\frac{2}{3}\right)$

    $= \frac{2}{21} + \frac{8}{21} = \frac{10}{21}$

    4. Now compute the numerator and the final probability …

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11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The probability that exactly 3 heads appear in six tosses of an unbiased coin, given that the first three tosses resulted in 2 or more heads is (A) $\dfrac{3}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{9}{16}$

    ›Reveal solutionSolution

    We use conditional probability: the desired probability is the ratio of the probability that exactly 3 heads occur AND the first three tosses have ≥2 heads, divided by the probability that the first three tosses have ≥2 heads. The result simplifies to $\frac{5}{16}$, so option (B) is correct.

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    Concept & Intuition

    The problem asks for a conditional probability:

    $P(\text{exactly 3 heads in 6 tosses} \mid \text{first 3 tosses have } \ge 2 \text{ heads})$

    We can’t just count all 6-toss sequences because the condition restricts the first three tosses. The key is to break the 6 tosses into two independent blocks of 3 tosses each (since coin tosses are independent). Then we count favorable outcomes in the first block (≥2 heads) and combine with outcomes in the second block that make the total exactly 3 heads.

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    Step-by-step solution

    1. Define events

       Let $A$ = “exactly 3 heads in 6 tosses”

       Let $B$ = “first 3 tosses have 2 or more heads”

       We want $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.

    2. Compute $P(B)$ — probability first 3 tosses have ≥2 heads.

       In 3 tosses of a fair coin, number of heads $X \sim \text{Binomial}(3, 1/2)$.

    $P(X \ge 2) = P(X=2) + P(X=3) = \binom{3}{2}\left(\frac12\right)^3 + \binom{3}{3}\left(\frac12\right)^3 = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac12.$

    3. Compute $P(A \cap B)$ — probability that exactly 3 heads total AND first 3 tosses have ≥2 heads.

       Let $h_1$ = heads in first 3 tosses, $h_2$ = heads in last 3 tosses.

       Total heads = $h_1 + h_2 = 3$, with $h_1 \ge 2$.

       Possible pairs $(h_1, h_2)$:

       • $h_1 = 2$, then $h_2 = 1$
       • $h_1 = 3$, then $h_2 = 0$

       Since the two blocks are independent:

    $P(h_1=2 \text{ and } h_2=1) = \left[\binom{3}{2}\left(\frac12\right)^3\right] \times \left[\binom{3}{1}\left(\frac12\right)^3\right] = \frac{3}{8} \cdot \frac{3}{8} = \frac{9}{64}$

    $P(h_1=3 \text{ and } h_2=0) = \left[\binom{3}{3}\left(\frac12\right)^3\right] \times \left[\binom{3}{0}\left(\frac12\right)^3\right] = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64}$

    So

    $$$$

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12. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.A dealer bought a fixed number of laptops from 3 different companies A, B and C. Among these laptops 28% are bought from A, 32% are bought from B and 40% are bought from C. 2.5% of laptops bought from A, 1.5% bought from B and 1% bought from C are likely to be defective. If a customer found that the laptop bought by him is defective, then the probability that it was from B is (A) $\dfrac{12}{39}$ (B) $\dfrac{49}{158}$ (C) $\dfrac{17}{78}$ (D) $\dfrac{24}{79}$

    ›Reveal solutionSolution

    This is a classic Bayes’ theorem problem: we are given the prior probabilities (percentages of laptops from each company) and the likelihoods (defect rates per company), and we need the posterior probability that a defective laptop came from B. The answer simplifies to $\frac{24}{79}$, which is option (D).

    We want $P(\text{B} \mid \text{Defective})$.

    The key idea: Bayes’ theorem lets us “reverse” conditional probabilities when we know the overall structure. Here, the “prior” is the proportion of laptops from each company, and the “likelihood” is the defect rate within each company. The “evidence” (total defect probability) is the weighted average of those defect rates.

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    Step-by-step reasoning

    1. Interpret the given percentages as probabilities.

       Let the total number of laptops be 100 (or any convenient number — the ratios are what matter).

       • $P(A) = 0.28$, $P(B) = 0.32$, $P(C) = 0.40$
       • Defect rates: $P(D \mid A) = 0.025$, $P(D \mid B) = 0.015$, $P(D \mid C) = 0.01$

    2. Find the total probability of a defective laptop.

       By the law of total probability:

    $P(D) = P(A)P(D \mid A) + P(B)P(D \mid B) + P(C)P(D \mid C)$

    Substitute:

    $P(D) = (0.28)(0.025) + (0.32)(0.015) + (0.40)(0.01)$

    Compute each term:

    • $0.28 \times 0.025 = 0.007$
    • $0.32 \times 0.015 = 0.0048$
    • $0.40 \times 0.01 = 0.004$ Sum: $0.007 + 0.0048 + 0.004 = 0.0158$

    3. Apply Bayes’ theorem.

    $P(B \mid D) = \frac{P(B) \, P(D \mid B)}{P(D)}$

    Substitute:

    $P(B \mid D) = \frac{0.32 \times 0.015}{0.0158} = \frac{0.0048}{0.0158}$

    4. Simplify the fraction. …

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