Q.Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2025Set 2025-A1 markMCQ
   Q.Three bags contain a number of red and white balls are as follows. Bag I: 3 red balls Bag II: 2 red balls and 1 white ball Bag III: 3 White balls The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}, i=1,2,3$. If a white ball is selected, what is the probablity that it came from Bag III (A) $\frac{9}{11}$ (B) $\frac{2}{11}$ (C) $0$ (D) $\frac{1}{11}$

   ›Reveal solutionSolution

   This is a classic Bayes’ theorem problem: we are given prior probabilities for choosing each bag and the conditional probabilities of drawing a white ball from each bag; after observing a white ball, we want the posterior probability that it came from Bag III. The answer is $\frac{9}{11}$.

   We start by noting that the problem gives us three bags with different compositions, and a rule for choosing a bag: the probability of picking bag $i$ is $\frac{i}{6}$. Then, from that bag, we pick a ball uniformly at random. We are told that a white ball was selected, and we need the probability that it came from Bag III. This is a textbook case for Bayes’ theorem, which lets us “reverse” conditional probabilities: we know $P(\text{white} \mid \text{bag } i)$, and we want $P(\text{bag } i \mid \text{white})$.

   Let’s work through it step by step.

   1. Define events and write down given probabilities. Let $B_i$ be the event that bag $i$ is chosen ($i=1,2,3$). We are told:

   $P(B_1) = \frac{1}{6},\quad P(B_2) = \frac{2}{6},\quad P(B_3) = \frac{3}{6}.$

   Let $W$ be the event that a white ball is drawn.

   From the bag compositions:

   • Bag I: 3 red, 0 white → $P(W \mid B_1) = 0$.
   • Bag II: 2 red, 1 white → $P(W \mid B_2) = \frac{1}{3}$.
   • Bag III: 0 red, 3 white → $P(W \mid B_3) = 1$.

   2. Compute the total probability of drawing a white ball. By the law of total probability:

   $P(W) = P(B_1)P(W\mid B_1) + P(B_2)P(W\mid B_2) + P(B_3)P(W\mid B_3).$

   Substitute:

   $P(W) = \frac{1}{6}\cdot 0 + \frac{2}{6}\cdot\frac{1}{3} + \frac{3}{6}\cdot 1 = 0 + \frac{2}{18} + \frac{3}{6} = \frac{1}{9} + \frac{1}{2}.$

   Get a common denominator (18):

   $P(W) = \frac{2}{18} + \frac{9}{18} = \frac{11}{18}.$

   3. Apply Bayes’ theorem to find $P(B_3 \mid W)$. Bayes’ theorem says:

   $P(B_3 \mid W) = \frac{P(B_3)\,P(W\mid B_3)}{P(W)}.$

   Plug in the numbers:

   $P(B_3 \mid W) = \frac{\frac{3}{6} \cdot 1}{\frac{11}{18}} = \frac{\frac{1}{2}}{\frac{11}{18}} = \frac{1}{2} \cdot \frac{18}{11} = \frac{9}{11}.$

   Tip

   Notice that Bag I contributes zero probability of a white ball, so it can be ignored entirely in the numerator — it only affects the denominator by adding zero. This often simplifies Bayes’ calculations: only bags that can produce the observed outcome matter.

   Watch out

   A common mistake is to forget that the prior probabilities $\frac{i}{6}$ are not equal — they favor higher-numbered bags. If you mistakenly treated all bags as equally likely, you would get $\frac{1}{2}$ instead of $\frac{9}{11}$.

   Thus, given that a white ball was drawn, the probability it came from Bag III is $\frac{9}{11}$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. 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):

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

   $P(\text{white}\mid T)=\tfrac{3}{10}+\tfrac{6}{10}\cdot\tfrac23+\tfrac{1}{10}\cdot\tfrac13=\tfrac{3}{10}+\tfrac{4}{10}+\tfrac{1}{30}=\tfrac{22}{30}=\tfrac{11}{15}.$

   With $P(H)=P(T)=\tfrac12$:

   $P(H\mid\text{white})=\frac{\tfrac12\cdot\tfrac45}{\tfrac12\cdot\tfrac45+\tfrac12\cdot\tfrac{11}{15}}=\frac{\tfrac45}{\tfrac45+\tfrac{11}{15}}=\frac{\tfrac{12}{15}}{\tfrac{23}{15}}=\frac{12}{23}.$

   ✓Final answer

   The correct option is (B) — $\frac{12}{23}$

3. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$.
   • First ball black ($P=\tfrac12$): urn now 5R, 7B ⇒ $P(\text{red})=\tfrac{5}{12}$.

   Total probability:

   $\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7+5}{24}=\frac{12}{24}=\frac12.$

   ✓Final answer

   The correct option is (B) — 1/2

4. 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.

   ---

   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.

   ---

   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.

   4. Identify which of these have the other side black.

      Only the red side $R_3$ (from card C) has a black other side. The other two red sides ($R_1, R_2$) have red on the other side.

   5. Compute the conditional probability.

   $P(\text{other side black} \mid \text{top side red}) = \frac{\text{number of favorable red sides}}{\text{total number of red sides}} = \frac{1}{3}.$

   Watch out

   The common mistake is to forget that the all-red card contributes two red sides to the sample space, making it twice as likely as the mixed card once we condition on seeing red.

   Tip

   A neat way to think: "Probability = (number of red sides with black opposite) / (total number of red sides)." This avoids the card-counting trap entirely.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

   ›Reveal solutionSolution

   Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

   Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

   Case 1 — first ball is green. $P = \dfrac{2}{7}$.

   Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

   $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

   Case 2 — first ball is red. $P = \dfrac{5}{7}$.

   Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$).

   $P(\text{2nd red}\mid \text{red first}) = \frac{4}{7}$

   Total probability:

   $P(\text{2nd red}) = \frac{2}{7}\cdot\frac{6}{7} + \frac{5}{7}\cdot\frac{4}{7} = \frac{12}{49} + \frac{20}{49} = \frac{32}{49}$

   ✓Final answer

   $P(\text{second ball red}) = \dfrac{32}{49}$ — option (B).

6. KCET 2021Set A-11 markMCQ
   Q.A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30% of cars. 80% of cars at plant X and 90% of cars at plant Y are rated as standard quality. A car is chosen at random and is found to be of standard quality. The probability that it has come from plant X is (A) $\dfrac{56}{73}$ (B) $\dfrac{56}{84}$ (C) $\dfrac{56}{83}$ (D) $\dfrac{56}{79}$

   ›Reveal solutionSolution

   This is a reverse-probability question (effect → cause), so use Bayes' theorem with the total probability of a standard-quality car in the denominator.

   Step 1 — Define the events.

   • $X$: the car came from plant X — $P(X) = 0.70$
   • $Y$: the car came from plant Y — $P(Y) = 0.30$
   • $S$: the car is of standard quality

   The conditional (likelihood) data given:

   $P(S\mid X) = 0.80, \qquad P(S\mid Y) = 0.90$

   Note $X$ and $Y$ are mutually exclusive and exhaustive ($0.7 + 0.3 = 1$), which is exactly what Bayes' theorem needs.

   Step 2 — Why Bayes and not simple conditioning.

   We are told the effect (the chosen car is standard) and asked for the probability of the cause (it came from X). That inversion — $P(X\mid S)$ from $P(S\mid X)$ — is precisely Bayes' theorem:

   $P(X \mid S) = \frac{P(X)\,P(S\mid X)}{P(X)\,P(S\mid X) + P(Y)\,P(S\mid Y)}$

   Step 3 — Compute the numerator.

   $P(X)P(S\mid X) = 0.70 \times 0.80 = 0.56$

   Step 4 — Compute the denominator (total probability of a standard car).

   $P(S) = 0.70\times0.80 + 0.30\times0.90 = 0.56 + 0.27 = 0.83$

   Step 5 — Divide.

   $P(X\mid S) = \frac{0.56}{0.83} = \frac{56}{83}$

   Step 6 — Sanity check. The posterior $56/83 \approx 0.675$ is a little below the prior $0.70$ — which makes sense, because plant Y has the higher standard-quality rate ($90\% > 80\%$), so learning the car is standard shifts a little belief towards Y. ✓

   (The distractors 73, 84, 79 are what you get by mis-adding $0.56 + 0.27$; the denominator must be the total probability $0.83$.)

   ✓Final answer

   The correct option is (C) $\frac{56}{83}$.

   ANSWER: C

7. 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.

   LCM of 15 and 35 is 105:

   $\frac{2}{15}=\frac{14}{105},\qquad \frac{6}{35}=\frac{18}{105},$

   $P(F)=\frac{14}{105}+\frac{18}{105}=\frac{32}{105}.$

   Step 5 — Apply Bayes.

   $P(B\mid F)=\frac{18/105}{32/105}=\frac{18}{32}=\frac{9}{16}.$

   (Check: $P(A\mid F)=\frac{14}{32}=\frac{7}{16}$, and $\frac{9}{16}+\frac{7}{16}=1$ ✓. Note option (A) $\frac{7}{16}$ is the trap — it is the probability for temple A.)

   ✓Final answer

   The correct option is (D) — $\dfrac{9}{16}$.

   ANSWER: D

8. 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):

   $P(B\mid E)=\frac{P(E\mid B)}{P(E\mid A)+P(E\mid B)}=\frac{\frac{24}{49}}{\frac{3}{5}+\frac{24}{49}}.$

   Combine the denominator over $245$:

   $\frac{3}{5}+\frac{24}{49}=\frac{147+120}{245}=\frac{267}{245}.$

   $P(B\mid E)=\frac{24}{49}\cdot\frac{245}{267}=\frac{24\cdot 5}{267}=\frac{120}{267}=\frac{40}{89}.$

   ✓Final answer

   $P(B\mid E)=\dfrac{40}{89}$, which is option (C).

9. COMEDK 2024Set 2024-M1 markMCQ
   Q.An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is (A) $\frac{17}{30}$ (B) $\frac{37}{60}$ (C) $\frac{31}{60}$ (D) $\frac{23}{30}$

   ›Reveal solutionSolution

   Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

   Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$, total $-1$). Drawing black puts it back and adds one more black (blacks $+1$, total $+1$).

   First two draws and the composition just before the 3rd draw

   • WW: $\tfrac24\cdot\tfrac13 = \tfrac16$; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$.
   • WB: $\tfrac24\cdot\tfrac23 = \tfrac13$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BW: $\tfrac24\cdot\tfrac25 = \tfrac15$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$.

   (The four path probabilities sum to $\tfrac16+\tfrac13+\tfrac15+\tfrac{3}{10} = 1$.)

   Total probability

   $P(\text{3rd black}) = \tfrac16(1) + \tfrac13\cdot\tfrac34 + \tfrac15\cdot\tfrac34 + \tfrac{3}{10}\cdot\tfrac23$

   $= \frac{10}{60} + \frac{15}{60} + \frac{9}{60} + \frac{12}{60} = \frac{46}{60} = \frac{23}{30}$

   ✓Final answer

   $P(\text{third ball is black}) = \dfrac{23}{30}$ — option (D).

   ANSWER: D

10. COMEDK 2026Set 2026-M1 markMCQ
    Q.Advika chooses one of three scarves every morning: Red, Blue, or Green. The probability she chooses Red is $20 \%$. The probability she chooses Blue is twice the probability of choosing Red. On the remaining days she wears a Green scarf. Once a scarf is chosen, she decides whether to wear a Hat (H) and Sunglasses (S). These choices are independent of each other but depend on the scarf colour: $\begin{array}{|l|l|l|} \hline \text { Scarf colour } & \mathbf{P ( H )} & \mathbf{P ( S )} \\ \hline \text { Red } & 0.5 & 0.8 \\ \hline \text { Blue } & 0.4 & 0.5 \\ \hline \text { Green } & 0.1 & 0.5 \\ \hline \end{array}$ Advika is spotted outdoors wearing both a Hat and Sunglasses. What is the probability that she is wearing the Red scarf? (A) $\frac{13}{313}$ (B) $\frac{8}{21}$ (C) $\frac{4}{9}$ (D) $\frac{8}{13}$

    ›Reveal solutionSolution

    Bayes' theorem on scarf colour given that both a hat and sunglasses are worn. Priors $P(R)=0.2,\ P(B)=0.4,\ P(G)=0.4$; likelihoods $P(H\cap S\mid\text{colour})=P(H)\,P(S)$. The posterior $P(R\mid H\cap S)=\frac{4}{9}$ — option (C).

    Concept. Hat and sunglasses are independent given the scarf, so $P(H\cap S\mid\text{colour})=P(H\mid\text{colour})\cdot P(S\mid\text{colour})$. Bayes' theorem then reverses the conditioning to give the probability of the scarf colour from the observed accessories.

    Step 1 — Priors.

    $P(R)=20\%=0.2,\qquad P(B)=2P(R)=0.4,\qquad P(G)=1-0.2-0.4=0.4.$

    Step 2 — Likelihood of wearing both accessories for each colour.

    $P(H\cap S\mid R)=0.5\times0.8=0.40,$

    $P(H\cap S\mid B)=0.4\times0.5=0.20,$

    $P(H\cap S\mid G)=0.1\times0.5=0.05.$

    Step 3 — Total probability of both accessories (denominator).

    $P(H\cap S)=(0.2)(0.40)+(0.4)(0.20)+(0.4)(0.05)=0.08+0.08+0.02=0.18.$

    Step 4 — Posterior for Red.

    $P(R\mid H\cap S)=\frac{P(R)\,P(H\cap S\mid R)}{P(H\cap S)}=\frac{0.08}{0.18}=\frac{8}{18}=\frac{4}{9}.$

    ✓Final answer

    $P(\text{Red}\mid H\cap S)=\dfrac{4}{9}$ — option (C).

    ANSWER: C

11. 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.

       From the list above, there are exactly 2 outcomes: $(1,2)$ and $(2,1)$.

    4. Apply conditional probability.

       The probability that the sum is 3, given that the sum is less than 6, is:

    $P(\text{sum}=3 \mid \text{sum}<6) = \frac{\text{Number of outcomes with sum}=3 \text{ and sum}<6}{\text{Number of outcomes with sum}<6}$

    Since every outcome with sum = 3 automatically satisfies sum < 6, the numerator is just 2. The denominator is 10.

    So:

    $P = \frac{2}{10} = \frac{1}{5}$

    Tip

    You can also think of this as: "Out of the 10 equally likely ways to get a sum less than 6, exactly 2 give a sum of 3." That’s the same as $\frac{2}{10}$.

    ✓Final answer

    The correct option is (C) $\dfrac{1}{5}$.

12. 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.

    First find $P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{1/6}{1/2} = \dfrac{1}{3}$. Since $A$ and $A'$ partition the space, $P(A|B) + P(A'|B) = 1$, so

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

    Both routes agree. (Note $P(A) = \tfrac13$ was not even needed — a useful reminder that a conditional probability given $B$ depends only on how $B$ is split.)

    ✓Final answer

    The correct option is (A) — $\dfrac{2}{3}$.

    ANSWER: A