Q.In Exercise 64 above, $P(B \mid A')$ is equal to
(A) $\dfrac{1}{5}$
(B) $\dfrac{3}{10}$
(C) $\dfrac{1}{2}$
(D) $\dfrac{3}{5}$

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2025Set 2025-A1 markMCQ
   Q.If for two events $A$ and $B, P(A-B)=\frac{1}{5}$ and $P(A)=\frac{3}{5}$ then $P(B / A)=$ (A) $\frac{2}{3}$ (B) $\frac{1}{2}$ (C) $\frac{3}{5}$ (D) $\frac{2}{5}$

   ›Reveal solutionSolution

   The key is to interpret $P(A-B)$ as $P(A \cap B^c)$ and use the definition of conditional probability. The result is $P(B/A) = \frac{2}{3}$, so option (A) is correct.

   We are asked for $P(B/A)$, the probability of $B$ given $A$. The definition is

   $P(B/A) = \frac{P(A \cap B)}{P(A)}.$

   We know $P(A) = \frac{3}{5}$, so we need $P(A \cap B)$. The given $P(A-B) = \frac{1}{5}$ is the key: $A-B$ means “$A$ and not $B$,” i.e., $A \cap B^c$.

   1. Relate $P(A-B)$ to $P(A \cap B)$ Since $A$ is the union of the disjoint parts “$A$ and $B$” and “$A$ and not $B$,” we have

   $P(A) = P(A \cap B) + P(A \cap B^c).$

   Here $P(A \cap B^c) = P(A-B) = \frac{1}{5}$ and $P(A) = \frac{3}{5}$.

   2. Solve for $P(A \cap B)$

   $\frac{3}{5} = P(A \cap B) + \frac{1}{5} \quad\Rightarrow\quad P(A \cap B) = \frac{3}{5} - \frac{1}{5} = \frac{2}{5}.$

   3. Apply the conditional probability formula $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{2/5}{3/5} = \frac{2}{3}.$ …

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2. KCET 2021Set A-11 markMCQ
   Q.Given that A and B are two events such that $P(B) = \dfrac{3}{5}$, $P(A/B) = \dfrac{1}{2}$ and $P(A \cup B) = \dfrac{4}{5}$, then $P(A) =$ (A) $\dfrac{3}{10}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Turn the conditional probability into $P(A\cap B)$, then substitute into the addition theorem and solve for $P(A)$.

   Step 1 — Extract $P(A\cap B)$ from the conditional probability.

   By definition,

   $P(A/B)=\frac{P(A\cap B)}{P(B)}\qquad (P(B)\neq0).$

   Rearranging (the multiplication theorem):

   $P(A\cap B)=P(A/B)\cdot P(B)=\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}.$

   This is the key move — the conditional probability is not directly usable in the union formula, but $P(A\cap B)$ is.

   Step 2 — Apply the addition theorem.

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

   The intersection is subtracted because the elements common to $A$ and $B$ would otherwise be counted twice.

   Step 3 — Substitute the known values.

   $\frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}.$

   Step 4 — Solve for $P(A)$. …

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3. KCET 2022Set C-41 markMCQ
   Q.If A and B are two events such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$ and $P(A|B) = \dfrac{1}{4}$, then $P(A' \cap B')$ is (A) $3/16$ (B) $1/12$ (C) $3/4$ (D) $1/4$

   ›Reveal solutionSolution

   Get $P(A\cap B)$ from the conditional probability, use the addition rule for $P(A\cup B)$, then apply De Morgan: $P(A'\cap B')=1-P(A\cup B)$.

   Step 1 — Recover the joint probability from the conditional.

   Conditional probability is defined by $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ — it re-scales the probability of $A$ to the reduced sample space $B$. Rearranging gives the multiplication rule:

   $P(A\cap B)=P(A|B)\,P(B)=\frac14\times\frac13=\frac1{12}.$

   Step 2 — Addition rule for the union.

   $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac12+\frac13-\frac1{12}.$

   Taking LCM $12$: …

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4. KCET 2020Set A-11 markMCQ
   Q.If A and B are two events such that $P(A) = \frac{1}{3}$, $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{6}$, then $P(A'/B)$ is (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) $\frac{1}{12}$

   ›Reveal solutionSolution

   Use $P(A'|B) = 1 - P(A|B)$ (equivalently $\frac{P(B)-P(A\cap B)}{P(B)}$), which gives $\frac{2}{3}$.

   Step 1 — The definition of conditional probability.

   For $P(B) > 0$,

   $P(A'|B) = \frac{P(A' \cap B)}{P(B)}$

   Conditioning on $B$ means we restrict the sample space to $B$ and ask what fraction of $B$ lies outside $A$.

   Step 2 — Split $B$ into the part inside $A$ and the part outside $A$.

   The events $A \cap B$ and $A' \cap B$ are disjoint and their union is exactly $B$:

   $P(B) = P(A \cap B) + P(A' \cap B)$

   $\Rightarrow P(A' \cap B) = P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$

   Step 3 — Divide by $P(B)$.

   $P(A'|B) = \frac{P(A' \cap B)}{P(B)} = \frac{\tfrac{1}{3}}{\tfrac{1}{2}} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}$

   Step 4 — Cross-check with the complement rule. …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.If A and B are two events such that $P(\bar{A})=0.3, P(B)=0.4, P(A \cap \bar{B})=0.5$, then find the value of $P(B / A \cup \bar{B})$ (A) 0.33 (B) 0.7 (C) 0.8 (D) 0.25

   ›Reveal solutionSolution

   The problem asks for a conditional probability $P(B \mid A \cup \bar{B})$. Using the definition of conditional probability and the given probabilities, we compute $P(B \cap (A \cup \bar{B}))$ and $P(A \cup \bar{B})$, then divide. The final value is $0.25$, so the correct option is (D).

   We are given:

   • $P(\bar{A}) = 0.3$, so $P(A) = 1 - 0.3 = 0.7$.
   • $P(B) = 0.4$.
   • $P(A \cap \bar{B}) = 0.5$.

   We need $P(B \mid A \cup \bar{B})$, which is the probability that $B$ occurs given that $A$ or $\bar{B}$ (or both) occurs.

   Concept and intuition:

   Conditional probability $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$. Here $X = B$ and $Y = A \cup \bar{B}$. So we need the probability that both $B$ and $(A \cup \bar{B})$ happen, divided by the probability that $A \cup \bar{B}$ happens. The key is to simplify the intersection $B \cap (A \cup \bar{B})$ using set algebra — it often collapses to something simpler.

   1. Simplify the numerator $P(B \cap (A \cup \bar{B}))$

      Using the distributive law:

      $B \cap (A \cup \bar{B}) = (B \cap A) \cup (B \cap \bar{B})$

      But $B \cap \bar{B} = \emptyset$, so this is just $B \cap A$.

      Hence $P(B \cap (A \cup \bar{B})) = P(A \cap B)$.

   2. Find $P(A \cap B)$

      We know $P(A) = 0.7$ and $P(A \cap \bar{B}) = 0.5$.

      Since $A = (A \cap B) \cup (A \cap \bar{B})$ and these are disjoint,

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

      $\Rightarrow 0.7 = P(A \cap B) + 0.5$

      $\Rightarrow P(A \cap B) = 0.2$.

   3. Find the denominator $P(A \cup \bar{B})$

      Use the inclusion-exclusion principle: …

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6. COMEDK 2023Set 2023-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } P(B)=\frac{3}{5} \quad P(A / B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5} \text { then } P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right)=$

   (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) 1 (D) $\frac{1}{5}$

   ›Reveal solutionSolution

   Step 5: sum = 1/5 + 4/5 = 1.

   Concept: conditional probability, addition theorem, complements, De Morgan.

   Given P(B) = 3/5, P(A|B) = 1/2, P(A U B) = 4/5.

   Step 1: P(A ^ B) = P(A|B) P(B) = (1/2)(3/5) = 3/10.

   Step 2: from P(A U B) = P(A) + P(B) - P(A ^ B),

   P(A) = 4/5 - 3/5 + 3/10 = 1/5 + 3/10 = 1/2.

   Step 3: P((A U B)') = 1 - P(A U B) = 1 - 4/5 = 1/5. …

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7. COMEDK 2024Set 2024-M1 markMCQ
   Q.A and B each have a calculator which can generate a single digit random number from the set $\{1,2,3,4,5,6,7,8\}$. They can generate a random number on their calculator. Given that the sum of the two numbers is 12 , then the probability that the two numbers are equal is (A) $\frac{5}{64}$ (B) $\frac{1}{5}$ (C) $\frac{1}{16}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   We are asked for the conditional probability that two numbers are equal given their sum is 12.

   The only equal pair summing to 12 is (6,6), and there are 5 total pairs summing to 12.

   So the probability is $\frac{1}{5}$, which corresponds to option (B).

   ---

   Concept and intuition

   This is a classic conditional probability problem: we are not interested in all possible outcomes, only those where the sum is exactly 12. The phrase “given that the sum is 12” means we restrict our universe to those pairs. Then we count how many of those restricted outcomes have the two numbers equal. The trap is to forget to restrict the denominator — many students mistakenly use the total number of all possible pairs (64) instead of only the favorable-sum pairs.

   ---

   Step-by-step solution

   1. Identify the sample space

      Each of A and B picks a digit from $\{1,2,\dots,8\}$.

      Total possible ordered pairs $(a,b)$: $8 \times 8 = 64$.

   2. List all pairs with sum 12

      We need $a + b = 12$, with $1 \le a,b \le 8$.

      Possible values for $a$:

      • If $a = 4$, then $b = 8$
      • If $a = 5$, then $b = 7$
      • If $a = 6$, then $b = 6$
      • If $a = 7$, then $b = 5$
      • If $a = 8$, then $b = 4$

      So the pairs are:

   $(4,8),\ (5,7),\ (6,6),\ (7,5),\ (8,4)$

   That’s 5 ordered pairs.

   3. Count the favorable outcomes “The two numbers are equal” means $a = b$. …

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.P and Q are considering to apply for a job. The probability that P applies for the job is $\frac{1}{4}$. The probability that $\mathrm{P}$ applies for the job given that $\mathrm{Q}$ applies for the job is $\frac{1}{2}$, and the probability that Q applies for the job given that P applies for the job is $\frac{1}{3}$. Then the probability that $\mathrm{P}$ does not apply for the job given that $\mathrm{Q}$ does not apply for the job is (A) $\frac{4}{5}$ (B) $\frac{7}{8}$ (C) $\frac{5}{6}$ (D) $\frac{11}{12}$

   ›Reveal solutionSolution

   We are given conditional probabilities and need to find $P(\overline{P} \mid \overline{Q})$. Using the definitions of conditional probability and the law of total probability, we compute $P(Q)$ and $P(\overline{Q})$, then apply Bayes' theorem to get $\frac{4}{5}$, which corresponds to option (A).

   Concept and intuition:

   This is a classic problem of working backwards from conditional probabilities to find a joint probability table. We know $P(P)$, $P(P \mid Q)$, and $P(Q \mid P)$. From these, we can find $P(P \cap Q)$ in two ways, which lets us solve for $P(Q)$. Then we can compute the desired conditional probability $P(\overline{P} \mid \overline{Q})$ using the complement rule and the definition of conditional probability.

   Step-by-step solution:

   1. Write down what is given. Let $P$ = event that P applies, $Q$ = event that Q applies. We have:

   $P(P) = \frac{1}{4}, \quad P(P \mid Q) = \frac{1}{2}, \quad P(Q \mid P) = \frac{1}{3}.$

   2. Use the definition of conditional probability to express $P(P \cap Q)$ in two ways. From $P(P \mid Q) = \frac{P(P \cap Q)}{P(Q)}$, we get

   $P(P \cap Q) = P(P \mid Q) \cdot P(Q) = \frac{1}{2} P(Q).$

   From $P(Q \mid P) = \frac{P(P \cap Q)}{P(P)}$, we get

   $P(P \cap Q) = P(Q \mid P) \cdot P(P) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$

   3. Equate the two expressions for $P(P \cap Q)$ to find $P(Q)$.

   $\frac{1}{2} P(Q) = \frac{1}{12} \quad \Rightarrow \quad P(Q) = \frac{1}{6}.$

   4. Find $P(\overline{Q})$ and $P(P \cap \overline{Q})$.

   $P(\overline{Q}) = 1 - P(Q) = 1 - \frac{1}{6} = \frac{5}{6}.$

   Also, $P(P \cap \overline{Q}) = P(P) - P(P \cap Q) = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}.$

   5. Find $P(\overline{P} \cap \overline{Q})$. Since $\overline{P} \cap \overline{Q}$ is the complement of $P \cup Q$, we can use:

   $P(\overline{P} \cap \overline{Q}) = 1 - P(P \cup Q).$

   First, $P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}.$ …

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9. KCET 2025Set A-11 markMCQ
   Q.Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is $\dfrac{2}{5}$. If she visits temple A, $\dfrac{1}{3}$ is the probability that she meets her friend, whereas it is $\dfrac{2}{7}$ if she visits temple B. Meera met her friend at one of the two temples. The probability that she met her at temple B is (A) $\dfrac{7}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{9}{16}$

   ›Reveal solutionSolution

   The friend has already been met (the effect); we want the probability of the cause (temple B) — that reversal of conditioning is exactly Bayes' theorem.

   Step 1 — Name the events.

   Let $A$ = "Meera visits temple A", $B$ = "Meera visits temple B", $F$ = "she meets her friend".

   She visits only one of the two temples, so $A$ and $B$ are mutually exclusive and exhaustive:

   $P(A)=\frac25\quad\Longrightarrow\quad P(B)=1-\frac25=\frac35 .$

   Given: $P(F\mid A)=\dfrac13$, $P(F\mid B)=\dfrac27$.

   Step 2 — Why Bayes.

   We are told the outcome ($F$ happened) and asked for the probability of a cause ($B$). Bayes' theorem inverts the conditioning:

   $P(B\mid F)=\frac{P(B)\,P(F\mid B)}{P(A)\,P(F\mid A)+P(B)\,P(F\mid B)} .$

   The denominator is $P(F)$ by the law of total probability — the friend can be met on either branch.

   Step 3 — Compute the two branch probabilities.

   $P(A\cap F)=\frac25\times\frac13=\frac{2}{15},\qquad P(B\cap F)=\frac35\times\frac27=\frac{6}{35} .$

   Step 4 — Total probability of meeting the friend. …

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10. KCET 2025Set A-11 markMCQ
    Q.If A and B are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $P(A|B) = \dfrac{P(B)}{P(A)}$ (B) $P(A|B) < P(A)$ (C) $P(A|B) \geq P(A)$ (D) $P(A) = P(B)$

    ›Reveal solutionSolution

    Use $A \subset B \Rightarrow A \cap B = A$, then note that dividing $P(A)$ by $P(B) \le 1$ cannot make it smaller.

    Step 1 — Simplify the intersection.

    If every element of $A$ lies in $B$, then $A \cap B = A$. Hence

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

    Step 2 — Compare with $P(A)$.

    Every probability satisfies $0 < P(B) \le 1$ (we are told $P(B) \ne 0$). Therefore $\dfrac{1}{P(B)} \ge 1$, and multiplying the non-negative number $P(A)$ by a factor $\ge 1$ gives

    $P(A\mid B) = P(A)\cdot\frac{1}{P(B)} \;\ge\; P(A).$

    Equality holds exactly when $P(B) = 1$ (or when $P(A) = 0$).

    Intuition: conditioning on $B$ throws away all outcomes outside $B$ — but none of $A$ lies outside $B$. So $A$'s share of the shrunken sample space can only grow.

    Step 3 — Eliminate. …

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11. KCET 2020Set A-11 markMCQ
    Q.Events $E_1$ and $E_2$ form a partition of the sample space S. A is any event such that $P(E_1) = P(E_2) = \frac{1}{2}$, $P(E_2/A) = \frac{1}{2}$ and $P(A/E_2) = \frac{2}{3}$, then $P(E_1/A)$ is (A) $\frac{1}{2}$ (B) $\frac{2}{3}$ (C) $1$ (D) $\frac{1}{4}$

    ›Reveal solutionSolution

    A partition's posterior probabilities must add to 1, so $P(E_1|A) = 1 - P(E_2|A) = 1 - \tfrac12 = \tfrac12$.

    Step 1 — What "partition of the sample space" means.

    $E_1$ and $E_2$ form a partition of $S$ if they are mutually exclusive ($E_1 \cap E_2 = \varnothing$) and exhaustive ($E_1 \cup E_2 = S$), with non-zero probabilities. So exactly one of them must occur.

    Step 2 — Conditioning preserves the partition.

    Conditioning on an event $A$ (with $P(A) > 0$) just renormalises probabilities inside $A$; it does not destroy the partition. Formally,

    $A = (A \cap E_1) \cup (A \cap E_2), \qquad (A\cap E_1)\cap(A\cap E_2) = \varnothing$

    Dividing by $P(A)$:

    $\frac{P(A\cap E_1)}{P(A)} + \frac{P(A\cap E_2)}{P(A)} = \frac{P(A)}{P(A)} = 1$

    which is precisely

    $P(E_1|A) + P(E_2|A) = 1$

    The posterior probabilities of a partition must still sum to 1 — a fact worth remembering, since it turns many Bayes questions into a one-line subtraction.

    Step 3 — Substitute the given value.

    $P(E_1|A) = 1 - P(E_2|A) = 1 - \frac{1}{2} = \frac{1}{2}$ …

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12. COMEDK 2023Set 2023-E1 markMCQ
    Q.A die is thrown twice and the sum of numbers appearing is observed to be 8 . What is the conditional probability that the number 5 has appeared atleast once? (A) $\frac{5}{36}$ (B) $\frac{2}{5}$ (C) $\frac{1}{18}$ (D) $\frac{1}{3}$

    ›Reveal solutionSolution

    Given the sum is $8$, the sample space is the $5$ ordered pairs summing to $8$; two of them include a $5$, giving probability $\frac{2}{5}$.

    The ordered outcomes with sum $8$ are

    $(2,6),(3,5),(4,4),(5,3),(6,2)\quad(5\text{ outcomes}).$ …

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