Q.An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$ $P(B \text{ fails alone}) = 0.15$ $P(A \text{ and B fail}) = 0.15$ Evaluate the following probabilities

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Let $A$ and $B$ be two events with $P(A) = \dfrac{1}{7}$, $P(A/B) = \dfrac{2}{5}$ and $P(B) = \dfrac{2}{7}$. Then the value $P(B/A)$ is (A) $\dfrac{1}{5}$ (B) $\dfrac{5}{49}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Chain the conditional-probability definition twice: first get $P(A\cap B)$ from $P(A/B)$, then get $P(B/A)$ from that.

   Concept and Intuition

   Conditional probability is defined as $P(E/F)=\dfrac{P(E\cap F)}{P(F)}$. Given one conditional probability, we can back out the joint probability $P(A\cap B)$, and then use it (with the definition again, but conditioning the other way) to find the other conditional probability.

   Step-by-Step Solution

   1. $P(A/B) = \dfrac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A/B)\cdot P(B) = \dfrac{2}{5}\times\dfrac{2}{7} = \dfrac{4}{35}$.
   2. Now use $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$. …

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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.2 aero planes I and II bond a target in succession. The probabilities of I and II scoring a hit correctly is 0.3 and 0.2 respectively. The second plane will bomb only if first misses the target. The probability that the target is hit by the $2^{nd}$ plane is (A) 0.06 (B) 0.14 (C) 0.32 (D) 0.7

   ›Reveal solutionSolution

   The 2nd plane gets a chance only after the 1st fails, so $P = 0.7\times 0.2 = 0.14$. Answer: (B).

   Concept and Intuition

   This is a sequential (conditional) experiment: the second trial happens only when the first fails. The event 'the target is hit by the 2nd plane' is therefore a compound event —

   $\{\text{I misses}\}\cap\{\text{II hits}\}$

   and because the two planes' performances are independent, the probability of the intersection is the product of the probabilities.

   A useful picture is a probability tree:

            ┌── I hits (0.3)  ────────────────► target hit by plane I  (0.3)
   Start ───┤
            └── I misses (0.7) ─┬── II hits (0.2) ──► hit by plane II  (0.7 × 0.2 = 0.14)
                                └── II misses (0.8) ► target not hit   (0.7 × 0.8 = 0.56)
   

   The three leaves sum to $0.3+0.14+0.56=1$ ✓ — a good check that the model is complete.

   Step-by-Step Solution

   1. Let $H_1$ = plane I hits, with $P(H_1)=0.3$, so $P(\overline{H_1})=1-0.3=0.7$.
   2. Let $H_2$ = plane II hits (given it bombs), with $P(H_2)=0.2$.
   3. Plane II bombs only if plane I missed. Hence

   $P(\text{target hit by 2nd plane}) = P(\overline{H_1}\cap H_2) = P(\overline{H_1})\,P(H_2)$

   4. Substitute: $= 0.7\times 0.2 = 0.14$ …

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3. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Two dice are rolled. If A denote the event that the same number shows on each die and B denote the event that the sum of the numbers on both dice is greater than 7, then $P(A|B)$ and $P(B|A)$ respectively are (A) $\dfrac{2}{5}, \dfrac{1}{4}$ (B) $\dfrac{1}{5}, \dfrac{1}{2}$ (C) $\dfrac{1}{5}, \dfrac{1}{4}$ (D) $\dfrac{1}{2}, \dfrac{3}{5}$

   ›Reveal solutionSolution

   Counting the 36 equally likely dice outcomes directly gives $P(A|B)=\frac15$ and $P(B|A)=\frac12$.

   Concept and Intuition

   With two fair dice there are $36$ equally likely outcomes, so every probability here reduces to simple counting: enumerate the outcomes in $A$, in $B$, and in $A\cap B$, then apply the conditional probability formula $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$ directly as a ratio of counts.

   Step-by-Step Solution

   1. $A$ = "same number on each die": outcomes $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ outcomes, so $P(A)=\dfrac6{36}$.
   2. $B$ = "sum $>7$", i.e. sum $\in\{8,9,10,11,12\}$. Counting pairs per sum: sum 8 has 5 pairs, 9 has 4, 10 has 3, 11 has 2, 12 has 1 — total $5+4+3+2+1=15$ outcomes, so $P(B)=\dfrac{15}{36}$.
   3. $A\cap B$: doubles with sum $>7$ — check each double: $(4,4)$ sum 8 ✓, $(5,5)$ sum 10 ✓, $(6,6)$ sum 12 ✓; $(1,1),(2,2),(3,3)$ have sums $2,4,6$, all $\le7$ — excluded. So $A\cap B$ has $3$ outcomes, $P(A\cap B)=\dfrac3{36}$. …

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4. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.In a toy factory, the machines A, B and C are used to manufacture 30%, 40% and 30% of the output, respectively. The probabilities of toys made by machines A, B, and C to be defective are respectively 2%, 3% and 1%. A toy is taken from the factory and is found to be defective. The probability that it was manufactured by the machine B is (A) $4/5$ (B) $2/9$ (C) $3/4$ (D) $4/7$

   ›Reveal solutionSolution

   This tests Bayes' theorem for finding the probability of a specific cause given an observed

   effect (a defective toy); the answer is $4/7$.

   Concept and Intuition

   When an outcome (a defective toy) could have come from several sources, each with its own prior

   probability and its own conditional probability of producing that outcome, Bayes' theorem lets us

   "invert" the conditioning: given that the outcome occurred, what's the probability it came from a

   particular source? The key is to first find the total probability of the outcome by summing over

   all sources (the law of total probability), then take the one source's contribution as a fraction of

   that total.

   Step-by-Step Solution

   1. Prior probabilities: $P(A)=0.3,\ P(B)=0.4,\ P(C)=0.3$.
   2. Conditional defect rates: $P(D\mid A)=0.02,\ P(D\mid B)=0.03,\ P(D\mid C)=0.01$.
   3. Total probability of a defective toy (law of total probability):

   $P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C) = 0.3(0.02)+0.4(0.03)+0.3(0.01)$

   $= 0.006+0.012+0.003 = 0.021.$

   4. By Bayes' theorem: …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In a sample space, $E$ is an event associated with the events $A$ and $B$. If $P(A)P(E|A) = l$ and $P(B)P(E|B) = m$ then, $P(B|E) =$ (A) $\dfrac{m}{l+m}$ always (B) $\dfrac{l}{l+m}$ only when $P(A)+P(B)=1$ (C) $\dfrac{m}{l+m}$ only when $P(A)+P(B)=1$ (D) $\dfrac{l}{l+m}$ always

   ›Reveal solutionSolution

   This tests when the "total probability" decomposition $P(E)=P(A\cap E)+P(B\cap E)$ is legitimate — only when $A,B$ partition the sample space. Answer: $\dfrac{m}{l+m}$, and only under that condition.

   Concept and Intuition

   Bayes' rule always gives $P(B\mid E) = P(B\cap E)/P(E)$. The numerator is handed to us as $m$. The tricky part is the denominator: $P(E)$ can only be written as $P(A\cap E)+P(B\cap E)=l+m$ if every outcome of $E$ passes through either $A$ or $B$ and not both — that is exactly the statement that $\{A,B\}$ partitions the sample space, equivalent to $A\cap B=\varnothing$ and $P(A)+P(B)=1$.

   Step-by-Step Solution

   1. From the given data, $P(A\cap E)=l$ and $P(B\cap E)=m$.
   2. By definition, $P(B\mid E)=\dfrac{P(B\cap E)}{P(E)}=\dfrac{m}{P(E)}$.
   3. If (and only if) $A,B$ partition the sample space, $E = (E\cap A)\cup(E\cap B)$ with no overlap, so $P(E)=l+m$.
   4. Substituting, $P(B\mid E)=\dfrac{m}{l+m}$ — but this substitution is valid only under the partition condition $P(A)+P(B)=1$. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.A, B, C are mutually exclusive and exhaustive events of a random experiment and E is an event that occurs in conjunction with one of the events A, B, C. The conditional Probabilities of E given the happening of A, B, C are respectively 0.6, 0.3 and 0.1. If $P(A) = 0.30$ and $P(B) = 0.50$, then $P(C|E) =$ (A) $\dfrac{2}{35}$ (B) $\dfrac{15}{35}$ (C) $\dfrac{18}{35}$ (D) $\dfrac{17}{35}$

   ›Reveal solutionSolution

   This is a direct Bayes' theorem application over three mutually exclusive, exhaustive causes; $P(C|E)=2/35$.

   Concept and Intuition

   When an event $E$ can occur alongside any of several mutually exclusive, exhaustive causes $A,B,C$, Bayes' theorem lets us find the probability of a particular cause given that $E$ has occurred, by weighing each cause's prior probability by how likely $E$ is under it, then normalizing.

   Step-by-Step Solution

   1. Since $A,B,C$ are mutually exclusive and exhaustive, $P(C)=1-P(A)-P(B)=1-0.30-0.50=0.20$.
   2. Total probability of $E$: $P(E)=P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)$ $=0.6(0.30)+0.3(0.50)+0.1(0.20)=0.18+0.15+0.02=0.35$ …

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7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.An item is tested on a device for its defectiveness. The probability that such an item is defective is 0.3. The device gives accurate result in 8 out of 10 such tests. If the device reports that an item tested is not defective, then the probability that it is actually defective is (A) $\dfrac{2}{15}$ (B) $\dfrac{3}{29}$ (C) $\dfrac{3}{31}$ (D) $\dfrac{4}{51}$

   ›Reveal solutionSolution

   A direct Bayes'-theorem inversion problem; the item is actually defective with probability $\dfrac{3}{31}$ given a "not defective" report — option (C).

   Concept and Intuition

   The device's report can be wrong. To find the true state given an observed (possibly wrong) report, we must weigh both ways the report could have arisen: a genuinely non-defective item correctly reported as such, or a genuinely defective item incorrectly reported as non-defective. Bayes' theorem combines these.

   Step-by-Step Solution

   1. Let $D$: item defective, $P(D)=0.3$; $D'$: item not defective, $P(D')=0.7$.
   2. Device accuracy $=0.8$, so it errs with probability $0.2$.
   3. "Reports not defective" while the item is defective means the device made an error: $P(\text{report ND}\mid D)=0.2$.
   4. "Reports not defective" while the item is not defective means the device was accurate: $P(\text{report ND}\mid D')=0.8$. …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a biased die, the probabilities for different faces to turn up are given below

   Face	1	2	3	4	5	6
   Probability	0.1	0.32	0.21	0.15	0.05	0.17

   The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is (A) $\dfrac{10}{33}$ (B) $\dfrac{5}{21}$ (C) $\dfrac{8}{21}$ (D) $\dfrac{1}{42}$

   ›Reveal solutionSolution

   Conditioning on "face 1 or face 2" just means renormalizing the two individual probabilities so they add to 1.

   Concept and Intuition

   $P(\text{face 1}\mid\text{face 1 or 2})=\dfrac{P(\text{face 1})}{P(\text{face 1})+P(\text{face 2})}$, since these two events are mutually exclusive and their union is the conditioning event.

   Step-by-Step Solution

   1. $P(1)=0.1$, $P(2)=0.32$.
   2. $P(1\text{ or }2)=0.1+0.32=0.42$.
   3. $P(1\mid1\text{ or }2)=\dfrac{0.1}{0.42}=\dfrac{10}{42}=\dfrac{5}{21}$.

   Common Mistakes …

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9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.It is given that in a random experiment events A and B are such that $P(A) = \dfrac{1}{4}$, $P(A|B) = \dfrac{1}{2}$ and $P(B|A) = \dfrac{2}{3}$ then $P(B) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Using $P(B|A)$ to find $P(A\cap B)$, then dividing by $P(A|B)$, gives $P(B)=\dfrac13$.

   Concept and Intuition

   The conditional probability definitions $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ and $P(B|A)=\dfrac{P(A\cap B)}{P(A)}$ share the common quantity $P(A\cap B)$ — computing it from one equation lets us solve the other for the unknown probability.

   Step-by-Step Solution

   1. From $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac23$, and $P(A)=\dfrac14$: $P(A\cap B)=\dfrac23\times\dfrac14=\dfrac16$.
   2. From $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac12$: $P(B)=\dfrac{P(A\cap B)}{P(A|B)}=\dfrac{1/6}{1/2}=\dfrac13$.

   Common Mistakes …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.If A,B are any two events of a random experiment and $P(B)\neq1$, then $P(A\mid B^c)=$ (A) $\dfrac{P(A)+P(A\cap B)}{1-P(B)}$ (B) $\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ (C) $\dfrac{P(A)+P(A\cap B)}{1+P(B)}$ (D) $\dfrac{P(A)}{1+P(B)}$

    ›Reveal solutionSolution

    Direct application of the conditional-probability definition to the complement event gives $P(A\mid B^c)=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$.

    Concept and Intuition

    $A\cap B^c$ is exactly the part of $A$ that does not overlap with $B$, so its probability is $P(A)$ minus the overlapping piece $P(A\cap B)$. Dividing by $P(B^c)=1-P(B)$ (valid since $P(B)\ne1$) gives the conditional probability.

    Step-by-Step Solution

    1. $P(A\mid B^c)=\dfrac{P(A\cap B^c)}{P(B^c)}$ by definition (needs $P(B^c)\ne0$, i.e. $P(B)\ne1$, as given).
    2. $A=(A\cap B)\cup(A\cap B^c)$, a disjoint union, so $P(A)=P(A\cap B)+P(A\cap B^c)\Rightarrow P(A\cap B^c)=P(A)-P(A\cap B)$.
    3. $P(B^c)=1-P(B)$. …

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11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.If A and B are two events such that $P(B) \neq 0$ and $P(B) \neq 1$, then $P(\bar{A}|\bar{B})$ is (A) $1-P(A|B)$ (B) $1-P(\bar{A}|B)$ (C) $\frac{1-P(A\cup B)}{P(\bar{B})}$ (D) $\frac{P(\bar{A})}{P(\bar{B})}$

    ›Reveal solutionSolution

    Directly apply the definition of conditional probability together with De Morgan's law; the answer is $\frac{1-P(A\cup B)}{P(\bar B)}$.

    Concept and Intuition

    Conditional probability is defined as $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$. Here $X=\bar A$, $Y=\bar B$. By De Morgan's law, $\bar A\cap\bar B = \overline{A\cup B}$, so its probability is $1-P(A\cup B)$.

    Step-by-Step Solution

    1. $P(\bar A|\bar B)=\dfrac{P(\bar A\cap\bar B)}{P(\bar B)}$.
    2. $\bar A\cap\bar B=\overline{A\cup B}$ (De Morgan).
    3. So $P(\bar A\cap\bar B)=1-P(A\cup B)$. …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.Two persons P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for the job is (A) $4/5$ (B) $5/6$ (C) $7/8$ (D) $11/12$

    ›Reveal solutionSolution

    Chain the given conditional probabilities to find $P(Q)$ and $P(P\cap Q)$, then use the complement rule — the answer is (A) $4/5$.

    Concept and Intuition

    Conditional probability definitions let us cross-multiply to recover the joint probability $P(P\cap Q)$ from either conditional. Once $P(P), P(Q), P(P\cap Q)$ are all known, De Morgan's law converts "neither event" into the complement of the union.

    Step-by-Step Solution

    1. Given: $P(P)=\tfrac14$, $P(P|Q)=\tfrac12$, $P(Q|P)=\tfrac13$.
    2. $P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$.
    3. Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$.
    4. $P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$.
    5. By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$: $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$.
    6. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$. …

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