Q.Determine $P(E|F)$. Two coins are tossed once, where

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1. 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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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 2023Set 2023-E1 markMCQ
   Q.Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is (A) $\frac{23}{30}$ (B) $\frac{12}{23}$ (C) $\frac{11}{23}$ (D) $\frac{19}{30}$

   ›Reveal solutionSolution

   Computing $P(\text{white drawn}\mid H)=\frac45$ and $P(\text{white}\mid T)=\frac{11}{15}$, Bayes' theorem gives $P(H\mid\text{white})=\frac{12}{23}$.

   Head (transfer 1 ball from A$(3W,2R)$ to B$(1W)$, then draw from B's 2 balls):

   $P(\text{white})=\tfrac35(1)+\tfrac25\left(\tfrac12\right)=\tfrac35+\tfrac15=\tfrac45.$

   Tail (transfer 2 balls from A to B$(1W)$, then draw from B's 3 balls):

   $P(WW)=\tfrac{3}{10}\Rightarrow P(\text{white})=1,\quad P(WR)=\tfrac{6}{10}\Rightarrow \tfrac23,\quad P(RR)=\tfrac{1}{10}\Rightarrow \tfrac13.$ …

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4. 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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5. 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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6. 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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7. COMEDK 2026Set 2026-A1 markMCQ
   Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

   ›Reveal solutionSolution

   $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

   Likelihoods of drawing one yellow and one green ($E$).

   Jar A (3 yellow, 2 green; two draws without replacement):

   $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

   Jar B (4 yellow, 3 green; two draws with replacement):

   $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

   Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels): …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

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   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

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   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities. …

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9. 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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10. COMEDK 2025Set 2025-A1 markMCQ
    Q.A bag contains $(n+1)$ coins. It is known that one of these coins has a head on both sides, whereas the other coins are fair. One of these coins is selected at random and tossed. If the probability that the toss results in heads is $\frac{7}{12}$, then the value of $n$ is : (A) 5 (B) 3 (C) 2 (D) 4

    ›Reveal solutionSolution

    The key idea is to treat the coin selection as a two‑case partition (two‑headed coin vs. fair coins) and apply the law of total probability. Solving the resulting equation gives $n = 5$, so the correct option is (A).

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    Concept and intuition

    We have a mixed bag: one trick coin that always lands heads, and $n$ fair coins that land heads with probability $\frac12$. When we pick a coin at random and toss it, the overall chance of heads is a weighted average of the two cases. The weight for the trick coin is $\frac{1}{n+1}$ (since there are $n+1$ coins total), and for any fair coin it’s $\frac{n}{n+1}$. The problem gives that overall probability as $\frac{7}{12}$, so we set up an equation and solve for $n$.

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    Step‑by‑step solution

    1. Define the events Let $T$ be the event that the two‑headed coin is selected, and $F$ the event that a fair coin is selected. Since selection is random,

    $P(T) = \frac{1}{n+1}, \qquad P(F) = \frac{n}{n+1}.$

    2. Conditional probabilities for heads If the trick coin is chosen, heads is certain:

    $P(\text{heads} \mid T) = 1.$

    If a fair coin is chosen, the chance of heads is $\frac12$:

    $P(\text{heads} \mid F) = \frac12.$

    3. Apply the law of total probability The overall probability of heads is

    $P(\text{heads}) = P(T) \cdot P(\text{heads}\mid T) + P(F) \cdot P(\text{heads}\mid F).$

    Substituting the values:

    $P(\text{heads}) = \frac{1}{n+1} \cdot 1 + \frac{n}{n+1} \cdot \frac12.$

    4. Simplify the expression

    $P(\text{heads}) = \frac{1}{n+1} + \frac{n}{2(n+1)} = \frac{2 + n}{2(n+1)}.$

    5. Set equal to the given probability The problem states this equals $\frac{7}{12}$: $$$$

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11. KCET 2025Set A-11 markMCQ
    Q.If A and B are two non-mutually exclusive events such that $P(A|B) = P(B|A)$, then (A) $A \subset B$ but $A \neq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$

    ›Reveal solutionSolution

    Write both conditional probabilities with the same numerator $P(A \cap B)$, cancel it (legal because the events are not mutually exclusive), and the denominators must be equal.

    Step 1 — Definition of conditional probability.

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

    Note $A \cap B = B \cap A$, so the two fractions share the same numerator.

    Step 2 — Impose the given condition.

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

    Step 3 — Cancel — and see why we are allowed to.

    The events are non-mutually-exclusive, i.e. $A \cap B \ne \phi$ and $P(A \cap B) \ne 0$. (This hypothesis is exactly what makes the cancellation valid — if $P(A\cap B)$ were 0, both sides would be 0 for any $P(A), P(B)$ and nothing would follow.) Dividing both sides by $P(A\cap B)$:

    $\frac{1}{P(B)} = \frac{1}{P(A)} \;\Longrightarrow\; P(A) = P(B).$

    Step 4 — Why the other options are not forced. …

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12. 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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