Q.Determine $P(E|F)$. A die is thrown three times, $E$ : 4 appears on the third toss, $F$ : 6 and 5 appears respectively on first two tosses.

Showing the 12 most recent of 45 on this concept.

1. KCET 2021Set A-11 markMCQ
   Q.Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is (A) $\dfrac{1}{18}$ (B) $\dfrac{5}{18}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   We use conditional probability: reduce the sample space to only those outcomes where the sum is less than 6, then find the fraction of those that give a sum of exactly 3. The answer is $\frac{1}{5}$.

   The key idea here is conditional probability — the probability of one event given that another event has already occurred. When we say "if it is known that the sum was less than 6", we are no longer considering all 36 possible outcomes of throwing two dice. Instead, we restrict ourselves to only those outcomes that satisfy the condition, and then ask: within this smaller set, what fraction gives a sum of 3?

   Let’s work it out step by step.

   1. List all possible outcomes when two dice are thrown.

      Each die shows a number from 1 to 6. The total number of ordered pairs $(a,b)$ is $6 \times 6 = 36$.

   2. Identify the condition: sum less than 6.

      The possible sums are 2, 3, 4, and 5 (since sum = 1 is impossible with two dice). Let’s list all ordered pairs that give each sum:

      • Sum = 2: $(1,1)$ → 1 outcome
      • Sum = 3: $(1,2), (2,1)$ → 2 outcomes
      • Sum = 4: $(1,3), (2,2), (3,1)$ → 3 outcomes
      • Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ → 4 outcomes

      Total outcomes with sum < 6: $1 + 2 + 3 + 4 = 10$.

      Watch out

      A common mistake is to forget that $(1,2)$ and $(2,1)$ are different outcomes. Dice are distinct, so order matters. Always count ordered pairs unless the problem explicitly says "identical dice" (which is rare in such problems).

   3. Identify the event of interest: sum = 3. …

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2. 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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3. COMEDK 2025Set 2025-M1 markMCQ
   Q.Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression? (A) $\frac{1}{8}$ (B) $\frac{1}{6}$ (C) $\frac{1}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The problem asks for the conditional probability that three dice sum to 15, given that the three numbers shown are distinct and form an arithmetic progression. By listing all such triples and counting those summing to 15, the probability is found to be 1/2.

   Concept and intuition

   We need $P(\text{sum}=15 \mid \text{distinct AP})$. The condition restricts the outcomes to only those triples $(a,b,c)$ with $a<b<c$ (since dice are fair, order matters but we can treat ordered rolls; however the condition “three different numbers” and “arithmetic progression” is symmetric). The key is to enumerate all possible ordered triples of distinct numbers from $\{1,2,3,4,5,6\}$ that are in arithmetic progression, then count how many of those sum to 15. Because the dice are fair, each ordered triple is equally likely, so the conditional probability is just the ratio of favorable ordered triples to total ordered triples satisfying the condition.

   Step-by-step reasoning

   1. Characterize arithmetic progressions of three distinct numbers from 1 to 6. Three numbers $x, y, z$ (with $x<y<z$) are in AP iff $2y = x+z$. Since they are distinct and from 1 to 6, the possible triples (unordered) are:

   $(1,2,3),\ (1,3,5),\ (2,3,4),\ (2,4,6),\ (3,4,5),\ (4,5,6).$

   Check: For (1,2,3): $2\cdot2=1+3$; (1,3,5): $2\cdot3=1+5$; etc. That’s all.

   2. Count ordered triples satisfying the condition.

      For each unordered triple, the three numbers are distinct, so they can appear in any order on the three dice. That gives $3! = 6$ permutations per triple.

      Total ordered triples satisfying the condition = $6 \times 6 = 36$.

   3. Identify which of these triples sum to 15.

      Compute sums of the unordered triples:

      $\begin{aligned} (1,2,3) &\to 6,\\ (1,3,5) &\to 9,\\ (2,3,4) &\to 9,\\ (2,4,6) &\to 12,\\ (3,4,5) &\to 12,\\ (4,5,6) &\to 15. \end{aligned}$ …

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4. KCET 2019Set A-11 markMCQ
   Q.A man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set $S = \{1, 2, 3, 4, 5, 6, 7\}$ and reports that it is even. The probability that it is actually even is (A) $\frac{2}{5}$ (B) $\frac{1}{5}$ (C) $\frac{1}{10}$ (D) $\frac{3}{5}$

   ›Reveal solutionSolution

   By Bayes' theorem the probability is $\frac{3}{5}$ — option (D).

   The set $S=\{1,2,3,4,5,6,7\}$ has $3$ even numbers $(2,4,6)$ and $4$ odd numbers, and one number is picked at random.

   Let $E$ = "the number is even" and $R$ = "the man reports it as even."

   Priors.

   $P(E)=\frac{3}{7},\qquad P(E^c)=\frac{4}{7}.$

   Likelihoods. He tells the truth with probability $\frac{2}{3}$ and lies with probability $\frac{1}{3}$.

   • If the number really is even, a truthful report says "even": $P(R\mid E)=\frac{2}{3}$.
   • If the number is odd, only a lie reports "even": $P(R\mid E^c)=\frac{1}{3}$.

   Bayes' theorem. …

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5. 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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6. COMEDK 2021Set 2021-B1 markMCQ
   Q.At a certain university 4% of male students are over 6 feet tall and 1% of female students are over 6 feet tall. The total student population is divided in the ratio 3 : 2, in favor of female students. If a student is selected at random from amongst all those over 6 feet tall, what is the probability that the student is a female? (A) 1/3 (B) 2/5 (C) 3/11 (D) 3/5

   ›Reveal solutionSolution

   Bayes' theorem gives $P(\text{female}\mid>6\text{ft})=\frac{0.006}{0.022}=\frac{3}{11}$.

   Population split $3:2$ in favour of females ⇒ $P(F)=\tfrac35=0.6,\ P(M)=\tfrac25=0.4$.

   Tall fractions: $P(T\mid F)=0.01,\ P(T\mid M)=0.04.$

   By Bayes' theorem: …

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7. 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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8. 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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9. 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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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.A coin is tossed until a head appears or until the coin has been tossed three times. Given that 'head' does not appear on the first toss, what is the probability that the coin is tossed thrice? (A) $\frac{1}{2}$ (B) $\frac{3}{8}$ (C) $\frac{1}{8}$ (D) $\frac{1}{4}$

    ›Reveal solutionSolution

    Given the first toss is a tail, the coin is tossed a third time only if the second toss is also a tail — probability $\tfrac12$.

    The coin stops the moment a head appears, or after three tosses. We are told the first toss is a tail (no head on toss 1).

    Now the second toss decides:

    • Head on toss 2 $\Rightarrow$ stop after 2 tosses.
    • Tail on toss 2 $\Rightarrow$ a third toss is made. …

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11. COMEDK 2026Set 2026-M1 markMCQ
    Q.Samhita faces a three-headed dragon. She wins a "Tactical medal" if she manages to defeat exactly one of the three heads. The battle proceeds head-by-head under the following conditions: The probability of defeating the first head is $\frac{\mathbf{1}}{\mathbf{3}}$. After a win: if she defeats a head, the probability of defeating the next head is $\frac{2}{3}$. After a loss: if she fails to defeat a head, the probability of defeating the next head is $\frac{\mathbf{1}}{\mathbf{4}}$. What is the probability that Samhita earns the "Tactical medal"? (A) $\frac{23}{72}$ (B) $\frac{5}{36}$ (C) $\frac{17}{72}$ (D) $\frac{19}{72}$

    ›Reveal solutionSolution

    Summing the three disjoint "exactly one win" paths gives $\dfrac{1}{12}+\dfrac{1}{18}+\dfrac{1}{8}=\dfrac{19}{72}$ — option (D).

    Set up the conditional probabilities. Let $W_i$ mean "defeats head $i$" and $L_i$ mean "fails":

    • $P(W_1) = \dfrac{1}{3}$, so $P(L_1) = \dfrac{2}{3}$.
    • After a win: next-head win probability $=\dfrac{2}{3}$, so next-head loss probability $=\dfrac{1}{3}$.
    • After a loss: next-head win probability $=\dfrac{1}{4}$, so next-head loss probability $=\dfrac{3}{4}$.

    Earning the medal means exactly one of the three heads is defeated. The three disjoint sequences are $W_1L_2L_3$, $L_1W_2L_3$, and $L_1L_2W_3$.

    Path 1 — $W_1 L_2 L_3$ (win, then loss after a win, then loss after a loss):

    $P = \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{36} = \frac{1}{12}$

    Path 2 — $L_1 W_2 L_3$ (loss, then win after a loss, then loss after a win): …

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12. 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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