Q.Two events $E$ and $F$ are independent. If $P(E) = 0.3$, $P(E \cup F) = 0.5$, then $P(E \mid F) - P(F \mid E)$ equals
(A) $\dfrac{2}{7}$
(B) $\dfrac{3}{35}$
(C) $\dfrac{1}{70}$
(D) $\dfrac{1}{7}$

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If E and F are events such that $P(\bar{F}) = 0.7$ and $P(E \cap F) = 0.2$, then $P(E|F)$ is (A) $2/3$ (B) $1/3$ (C) $3/4$ (D) $1/4$

   ›Reveal solutionSolution

   This tests the basic conditional-probability formula together with the complement rule. $P(E|F)=2/3$.

   Concept and Intuition

   $P(E|F)$ asks: given that $F$ has already happened, what fraction of that reduced sample space does $E\cap F$ occupy? It is always $\dfrac{P(E\cap F)}{P(F)}$, so first we need $P(F)$ itself, which is given indirectly through its complement.

   Step-by-Step Solution

   1. $P(\bar F) = 0.7 \Rightarrow P(F) = 1 - 0.7 = 0.3$.
   2. $P(E\cap F) = 0.2$ is given directly. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $E$: $P(E)=P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)$ $=0.6(0.30)+0.3(0.50)+0.1(0.20)=0.18+0.15+0.02=0.35$ …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Given $P(A) = 0.5, P(B) = 0.4, P(A \cap B) = 0.3$ then $P(A'/B')$ is equal to (A) $\dfrac{1}{3}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Using De Morgan's law $A'\cap B'=(A\cup B)'$ and the conditional probability formula gives $P(A'|B')=\dfrac23$.

   Concept and Intuition

   $P(A'|B')$ asks: given we're outside $B$, what's the chance we're also outside $A$? The key trick is that $A'\cap B' = (A\cup B)'$ (De Morgan), which is easy to compute from $P(A\cup B)$.

   Step-by-Step Solution

   1. $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.3=0.6$.
   2. $P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)=1-0.6=0.4$.
   3. $P(B')=1-P(B)=1-0.4=0.6$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If two events A and B are such that $P(\bar A)=0.3$, $P(B)=0.4$ and $P(A\cap\bar B)=0.5$, then $P(B/(A\cup\bar B))=$ (A) 0.25 (B) 0.6 (C) 0.45 (D) 0.8

   ›Reveal solutionSolution

   This tests careful set manipulation of conditional probability with a compound conditioning event $A\cup\bar B$. Answer: 0.25.

   Concept and Intuition

   Conditional probability $P(B\mid E) = \dfrac{P(B\cap E)}{P(E)}$ works exactly the same way when $E$ is a compound event like $A\cup\bar B$ — we just need to correctly compute $P(E)$ and $P(B\cap E)$ using set algebra (distributing intersection over union, and using the fact that $B$ and $\bar B$ are disjoint).

   Step-by-Step Solution

   1. From $P(\bar A)=0.3$, get $P(A)=1-0.3=0.7$.
   2. $P(A\cap\bar B)=0.5$ is given directly, so $P(A\cap B)=P(A)-P(A\cap\bar B)=0.7-0.5=0.2$.
   3. Compute $P(A\cup\bar B)$ using inclusion-exclusion: $P(A\cup\bar B)=P(A)+P(\bar B)-P(A\cap\bar B)$. Here $P(\bar B)=1-0.4=0.6$, so $P(A\cup\bar B)=0.7+0.6-0.5=0.8$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.There are three families $F_1, F_2, F_3$. $F_1$ has 2 boys and 1 girl; $F_2$ has 1 boy and 2 girls; $F_3$ has 1 boy and 1 girl. A family is randomly chosen and a child is chosen from that family randomly. If it is known that the child thus selected is a girl, then the probability that she is from $F_2$ is (A) $\dfrac{4}{9}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   Applying Bayes' theorem across the three equally-likely families gives $P(F_2\mid\text{girl})=\dfrac49$.

   Concept and Intuition

   This is a textbook application of Bayes' theorem: we're given the reverse conditional probabilities (family $\to$ probability of picking a girl) and asked for the forward one (girl picked $\to$ probability she's from a specific family).

   Step-by-Step Solution

   1. Prior: $P(F_1)=P(F_2)=P(F_3)=\dfrac13$.
   2. $P(\text{girl}\mid F_1)=\dfrac13$ (1 girl out of 3 children), $P(\text{girl}\mid F_2)=\dfrac23$, $P(\text{girl}\mid F_3)=\dfrac12$.
   3. Total probability: $P(\text{girl})=\frac13\cdot\frac13+\frac13\cdot\frac23+\frac13\cdot\frac12=\frac13\left(\frac13+\frac23+\frac12\right)=\frac13\cdot\frac{2+4+3}{6}=\frac13\cdot\frac96=\frac12$
   4. $P(F_2\cap\text{girl})=\dfrac13\cdot\dfrac23=\dfrac29$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.A die is thrown twice. Let A be the event of getting a prime number when the die is thrown first time and B be the event of getting an even number when the die is thrown second time. Then $P(A/\bar B)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Since $A$ and $B$ (hence $\bar B$) come from two independent throws of the die, conditioning on $\bar B$ does not change $P(A)$; the answer is $1/2$.

   Concept and Intuition

   When two events are determined by physically independent trials (first throw vs second throw of a die), any conditional probability between them collapses to the unconditional probability — conditioning on an independent event changes nothing.

   Step-by-Step Solution

   1. $A$: prime number on the first throw, i.e. outcome in $\{2,3,5\}$, so $P(A)=\dfrac{3}{6}=\dfrac{1}{2}$.
   2. $B$: even number on the second throw, i.e. outcome in $\{2,4,6\}$, so $P(B)=\dfrac{1}{2}$, and $P(\bar B)=\dfrac{1}{2}$ (odd on the second throw). …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.If $A$ and $B$ are mutually exclusive events with $P(A) = \dfrac{1}{4}$ and $P(B) = \dfrac{3}{7}$. Then what is the value of $P(A / A \cup B) =$ (A) $\dfrac{7}{19}$ (B) $\dfrac{12}{19}$ (C) $\dfrac{06}{19}$ (D) $\dfrac{13}{19}$

   ›Reveal solutionSolution

   Mutually exclusive events make $P(A\cup B)$ a simple sum, and since $A$ is entirely contained in $A\cup B$, the conditional probability reduces to $P(A)/P(A\cup B)$.

   Concept and Intuition

   For mutually exclusive $A,B$: $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$. Also $A\cap(A\cup B)=A$ always, so $P(A\mid A\cup B)=\dfrac{P(A\cap(A\cup B))}{P(A\cup B)}=\dfrac{P(A)}{P(A\cup B)}$.

   Step-by-Step Solution

   1. $P(A\cup B) = P(A)+P(B) = \dfrac14+\dfrac37$.
   2. Common denominator 28: $\dfrac{7}{28}+\dfrac{12}{28}=\dfrac{19}{28}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the probability that exactly 5 of the balls in the bag are green is: (A) $\dfrac{4}{35}$ (B) $\dfrac{5}{35}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{1}{7}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem: given a uniform prior on the unknown number of green balls in the bag, update the belief after observing 3 green balls drawn. Answer: $2/7$.

   Concept and Intuition

   Before drawing, any number of green balls from $0$ to $6$ is equally likely (7 equally likely hypotheses). Observing "all 3 drawn balls are green" is much more likely under hypotheses with more green balls, so Bayes' theorem reweights the prior toward higher $k$. We want the posterior probability that $k=5$.

   Step-by-Step Solution

   1. Prior: $P(k \text{ green balls})=\dfrac17$ for $k=0,1,\dots,6$.
   2. Likelihood of drawing 3 balls, all green, given $k$ green balls out of 6: $P(\text{all green}\mid k)=\dfrac{\binom{k}{3}}{\binom{6}{3}}=\dfrac{\binom{k}{3}}{20}$ (zero for $k<3$).
   3. Compute: $k=3:\binom33=1\Rightarrow1/20$; $k=4:\binom43=4\Rightarrow4/20$; $k=5:\binom53=10\Rightarrow10/20$; $k=6:\binom63=20\Rightarrow20/20=1$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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, $P(A\cap B)=\dfrac3{36}$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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: $P(\text{report ND}\mid D')=0.8$. …

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