Q.A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Showing the 12 most recent of 45 on this concept.

1. 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. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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.

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

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

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

   🔒 Start your 14-day free trial to see the full solution →
3. 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. …

   🔒 Start your 14-day free trial to see the full solution →
4. KCET 2020Set A-11 markMCQ
   Q.The probability of solving a problem by three persons A, B and C independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is (A) $\frac{1}{12}$ (B) $\frac{1}{4}$ (C) $\frac{1}{24}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   The problem asks for the probability that exactly two of the three persons solve it. We compute the sum of probabilities for each pair solving while the third fails, giving $\frac{1}{8}$.

   The key idea: "solved by any two of them" means exactly two solve it, not at least two. The third person must fail. Since A, B, and C work independently, we multiply their individual probabilities for each specific outcome and then add the three possible cases.

   A common mistake is to include the case where all three solve it. That would be "at least two", not "any two". The phrase "any two" in probability problems almost always means exactly two.

   Let’s denote:

   • $P(A) = \frac{1}{2}$, so $P(\text{A fails}) = 1 - \frac{1}{2} = \frac{1}{2}$
   • $P(B) = \frac{1}{4}$, so $P(\text{B fails}) = 1 - \frac{1}{4} = \frac{3}{4}$
   • $P(C) = \frac{1}{3}$, so $P(\text{C fails}) = 1 - \frac{1}{3} = \frac{2}{3}$

   We want exactly two successes. There are three mutually exclusive ways this happens:

   1. A and B solve, C fails

      Probability = $P(A) \times P(B) \times P(\text{C fails})$

      $= \frac{1}{2} \times \frac{1}{4} \times \frac{2}{3} = \frac{2}{24} = \frac{1}{12}$

   2. A and C solve, B fails

      Probability = $P(A) \times P(C) \times P(\text{B fails})$

      $= \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} = \frac{1}{8}$

   3. B and C solve, A fails

      Probability = $P(B) \times P(C) \times P(\text{A fails})$ …

   🔒 Start your 14-day free trial to see the full solution →
5. 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: …

   🔒 Start your 14-day free trial to see the full solution →
6. COMEDK 2025Set 2025-M1 markMCQ
   Q.In an entrance test, there are multiple choice questions. There are four possible answers to each question of which only one is correct. The probability that a student knows the answer to a question is $90 \%$. If he gets the correct answer to a question, then the probability that he was guessing is (A) $\frac{36}{37}$ (B) $\frac{1}{9}$ (C) $\frac{1}{37}$ (D) $\frac{37}{40}$

   ›Reveal solutionSolution

   We use Bayes’ theorem to reverse the conditional probability: given that the student answered correctly, we want the probability that he was guessing. The answer is $\frac{1}{37}$, so the correct option is (C).

   ---

   Concept and intuition

   This is a classic Bayes’ theorem problem. We have two ways a student can get a correct answer: either he knows the answer (and thus answers correctly for sure), or he guesses (and has a 1 in 4 chance of being correct). We are told the prior probability that he knows the answer is 90%. The question asks: given that he got it right, what’s the chance he was actually guessing? Bayes’ theorem lets us “flip” the conditional probability, using the known likelihoods.

   ---

   Step-by-step reasoning

   1. Define events clearly Let $K$ = “student knows the answer”, and $G$ = “student guesses” (so $G$ is the complement of $K$). Let $C$ = “student gives the correct answer”. We are given:

   $P(K) = 0.9, \quad P(G) = 0.1.$

   2. Conditional probabilities for answering correctly

      • If the student knows the answer, he is certain to be correct: $P(C \mid K) = 1$.
      • If he guesses, he picks randomly among 4 options, so $P(C \mid G) = \frac{1}{4}$.

   3. Find the total probability of a correct answer

      By the law of total probability:

   $P(C) = P(C \mid K)P(K) + P(C \mid G)P(G) = (1)(0.9) + \left(\frac{1}{4}\right)(0.1) = 0.9 + 0.025 = 0.925.$

   4. Apply Bayes’ theorem We want $P(G \mid C)$, the probability that he was guessing given that he answered correctly: $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. 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.

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

   🔒 Start your 14-day free trial to see the full solution →
8. 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:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{2/5}{3/5} = \frac{2}{3}.$ …

   🔒 Start your 14-day free trial to see the full solution →
10. 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):

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

    🔒 Start your 14-day free trial to see the full solution →
11. 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: …

    🔒 Start your 14-day free trial to see the full solution →
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. …

    🔒 Start your 14-day free trial to see the full solution →