Q.The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $A$ and $B$ are two events of a random experiment such that $P(A) = \dfrac{5}{6}$, $P(B) = \dfrac{3}{4}$ and $P(A \cap \bar{B}) = \dfrac{1}{6}$ then (A) $P(A \cup B) = 5[P(\bar{A} \cap B) + P(A \cap \bar{B})]$ (B) $2P(A \cup B) = 5P(A \cap B)$ (C) $3P(A \cup B) = 11P(A \cap \bar{B})$ (D) $P(A \cup B) = 11P(\bar{A} \cap B)$

   ›Reveal solutionSolution

   Tests basic set-algebra of probability (complement, union, intersection identities). Working out $P(A\cap B)$ from the given data pins down every other quantity, and testing each option shows only (D) holds.

   Concept and Intuition

   All of $P(A\cup B)$, $P(A\cap \bar B)$, $P(\bar A\cap B)$ can be written in terms of $P(A), P(B), P(A\cap B)$ using the basic set identities $A = (A\cap B)\cup(A\cap\bar B)$ (disjoint union) and the addition law. So once $P(A\cap B)$ is known everything else follows.

   Step-by-Step Solution

   1. Since $A\cap B$ and $A\cap \bar B$ partition $A$: $P(A) = P(A\cap B) + P(A\cap \bar B)$. $\dfrac{5}{6} = P(A\cap B) + \dfrac{1}{6} \Rightarrow P(A\cap B) = \dfrac{4}{6} = \dfrac{2}{3}$.
   2. Addition law: $P(A\cup B) = P(A)+P(B)-P(A\cap B) = \dfrac{5}{6}+\dfrac{3}{4}-\dfrac{2}{3}$. Common denominator 12: $\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{8}{12} = \dfrac{11}{12}$.
   3. Since $B\cap A$ and $\bar A\cap B$ partition $B$: $P(\bar A\cap B) = P(B) - P(A\cap B) = \dfrac{3}{4}-\dfrac{2}{3} = \dfrac{9}{12}-\dfrac{8}{12} = \dfrac{1}{12}$.
   4. Test option (D): $11P(\bar A\cap B) = 11\times\dfrac{1}{12} = \dfrac{11}{12} = P(A\cup B)$. This matches exactly. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = 0.8$, $P(E_2) = 0.7$ and $P(E_1 \cap E_2) \geq C$, then $C =$ (A) $0.5$ (B) $0.6$ (C) $0.7$ (D) $0.65$

   ›Reveal solutionSolution

   This is a direct application of Bonferroni's inequality $P(E_1\cap E_2)\ge P(E_1)+P(E_2)-1$. Plugging in the given probabilities gives $C=0.5$ — option (A).

   Concept and Intuition

   For any two events, $P(E_1\cup E_2)\le 1$ always. Since

   $P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2),$

   rearranging gives

   $P(E_1\cap E_2)\ge P(E_1)+P(E_2)-1.$

   This is Bonferroni's inequality — it gives the tightest universally-valid lower bound on the intersection probability purely from the individual probabilities, without knowing anything else about how $E_1,E_2$ overlap.

   Step-by-Step Solution

   1. We know $P(E_1\cup E_2)\le 1$ for any two events in the same sample space.
   2. Using the inclusion-exclusion identity: $P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)\le 1$.
   3. Rearranging: $P(E_1\cap E_2)\ge P(E_1)+P(E_2)-1$.
   4. Substitute the given values: $P(E_1\cap E_2)\ge 0.8+0.7-1 = 0.5$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.In a pack of 6 cards, two cards are marked with 5, two cards are marked with 6, one card is marked with 7 and another card is marked with 8. In another pack of 6 cards, one card is marked with 5, two cards are marked with 6, two cards are marked with 7 and one card is marked with 8. If one card from each pack is drawn at random, the probability that the sum of the numbers on the cards is 12 or 13 is (A) $\dfrac{3}{5}$ (B) $\dfrac{17}{36}$ (C) $\dfrac{7}{12}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   List every (Pack-1 value, Pack-2 value) pair summing to 12 or 13, weight each by how many cards carry those values, and divide by the total $36$ equally-likely draws.

   Concept and Intuition

   Since one card is drawn from each pack independently, every one of the $6\times6=36$ ordered pairs of cards is equally likely. The "value" on a card can repeat (e.g. two cards marked 5), so counting favourable cards (not just favourable values) means multiplying by how many cards carry each value.

   Step-by-Step Solution

   1. Pack 1 counts: value $5\to2$ cards, $6\to2$, $7\to1$, $8\to1$ (total 6).
   2. Pack 2 counts: value $5\to1$ card, $6\to2$, $7\to2$, $8\to1$ (total 6).
   3. Total equally-likely outcomes when drawing one from each: $6\times6=36$.
   4. Pairs $(v_1,v_2)$ with $v_1+v_2=12$: $(5,7)\to2\times2=4$; $(6,6)\to2\times2=4$; $(7,5)\to1\times1=1$. No $(8,4)$ since $4$ doesn't appear. Subtotal $=9$.
   5. Pairs with $v_1+v_2=13$: $(5,8)\to2\times1=2$; $(6,7)\to2\times2=4$; $(7,6)\to1\times2=2$; $(8,5)\to1\times1=1$. Subtotal $=9$. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If A and B are events of a random experiment such that $P(A\cup B) = \dfrac{3}{4}, P(A\cap B) = \dfrac{1}{4}, P(\overline{A}) = \dfrac{2}{3}$, then $P(\overline{A}\cap B) =$ (A) $\dfrac{5}{8}$ (B) $\dfrac{5}{12}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   $P(\overline A\cap B)$ is the part of $B$ that lies outside $A$, i.e. $P(B)-P(A\cap B)$. Working out $P(B)$ from the union formula gives the answer $5/12$.

   Concept and Intuition

   The event $\overline A\cap B$ is "$B$ occurs but $A$ doesn't" — geometrically, the part of set $B$ outside $A$. So $P(\overline A\cap B)=P(B)-P(A\cap B)$. We're given $P(A\cup B)$ and $P(A\cap B)$ directly, and $P(A)$ via its complement, so first find $P(B)$ using the addition rule.

   Step-by-Step Solution

   1. $P(\overline A)=2/3 \Rightarrow P(A)=1-2/3=1/3$.
   2. Addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
   3. $\dfrac34 = \dfrac13+P(B)-\dfrac14 \Rightarrow P(B)=\dfrac34-\dfrac13+\dfrac14=1-\dfrac13=\dfrac23$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A problem in Algebra is given to two students A and B whose chances of solving it are $\dfrac{2}{5}$ and $\dfrac{3}{4}$ respectively. The probability that the problem is solved if both of them try independently is (A) $\dfrac{17}{20}$ (B) $\dfrac{3}{20}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{13}{20}$

   ›Reveal solutionSolution

   With two independent solvers, it's easiest to compute the complement — the chance neither solves it — and subtract from 1, giving $\frac{17}{20}$.

   Concept and Intuition

   "At least one of two independent events occurs" is most cleanly computed via the complement rule: $P(\text{at least one})=1-P(\text{neither})$, and for independent events, $P(\text{neither})$ is just the product of each individual "fails" probability.

   Step-by-Step Solution

   1. $P(A \text{ solves})=\dfrac25$, so $P(A \text{ fails})=1-\dfrac25=\dfrac35$.
   2. $P(B \text{ solves})=\dfrac34$, so $P(B \text{ fails})=1-\dfrac34=\dfrac14$.
   3. Since A and B try independently, $P(\text{neither solves})=P(A\text{ fails})\times P(B\text{ fails})=\dfrac35\times\dfrac14=\dfrac{3}{20}$. …

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6. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If A is the event of getting a sum greater than 14 and B is the event of getting a sum which is a multiple of 3, then $P(A\cap\bar{B})+P(\bar{A}\cap B)=$ (A) $\dfrac{35}{108}$ (B) $\dfrac{17}{54}$ (C) $\dfrac{45}{108}$ (D) $\dfrac{5}{54}$

   ›Reveal solutionSolution

   This asks for the symmetric-difference probability of two events, which reduces to $P(A)+P(B)-2P(A\cap B)$; counting outcomes for three dice gives $\frac{35}{108}$.

   Concept and Intuition

   $P(A\cap\bar B)+P(\bar A\cap B)$ is the probability that exactly one of $A,B$ occurs (the symmetric difference $A\,\triangle\,B$). Since $P(A)=P(A\cap B)+P(A\cap\bar B)$ and $P(B)=P(A\cap B)+P(\bar A\cap B)$, adding and rearranging gives $P(A\cap\bar B)+P(\bar A\cap B)=P(A)+P(B)-2P(A\cap B)$ — so we just need the three probabilities $P(A)$, $P(B)$, $P(A\cap B)$, each obtainable by counting favourable outcomes among the $6^3=216$ equally likely triples.

   Step-by-Step Solution

   1. For three dice, the number of ordered triples summing to $s$ follows a known symmetric distribution around $s=10.5$: ways$(3)=1$, $(4)=3$, $(5)=6$, $(6)=10$, $(7)=15$, $(8)=21$, $(9)=25$, $(10)=27$, $(11)=27$, $(12)=25$, $(13)=21$, $(14)=15$, $(15)=10$, $(16)=6$, $(17)=3$, $(18)=1$ (total $=216$, confirming the count).
   2. $A$ = sum $>14$, i.e. sums $\{15,16,17,18\}$: ways $=10+6+3+1=20$. So $P(A)=\dfrac{20}{216}$.
   3. $B$ = sum is a multiple of 3, i.e. sums $\{3,6,9,12,15,18\}$: ways $=1+10+25+25+10+1=72$. So $P(B)=\dfrac{72}{216}=\dfrac13$.
   4. $A\cap B$ = sums in both sets = $\{15,18\}$: ways $=10+1=11$. So $P(A\cap B)=\dfrac{11}{216}$. …

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7. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The probability that a person A completes a work in a given time is $\dfrac{2}{3}$ and the probability that another person B completes the same work in the same time is $\dfrac{3}{4}$. If both A and B start doing this work at the same time, then the probability that the work is completed in the given time is (A) $\dfrac{11}{12}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{8}{9}$

   ›Reveal solutionSolution

   The work gets done if A or B (or both) finish it in time; use the complement of "both fail." Answer: $\dfrac{11}{12}$.

   Concept and Intuition

   "The work is completed in the given time" happens as long as at least one of the two independent workers finishes it — it does not require both to finish. The cleanest way to compute "at least one" probabilities is via the complement: $P(\text{at least one}) = 1 - P(\text{neither})$.

   Step-by-Step Solution

   1. Given: $P(A) = \dfrac23$, so $P(\bar A) = \dfrac13$.
   2. Given: $P(B) = \dfrac34$, so $P(\bar B) = \dfrac14$.
   3. Since A and B work independently, $P(\bar A \cap \bar B) = P(\bar A)\cdot P(\bar B) = \dfrac13\cdot\dfrac14 = \dfrac{1}{12}$.
   4. The work is completed unless both fail, so: $P(\text{completed}) = 1 - \frac{1}{12} = \frac{11}{12}$ …

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8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.There are 8 boys and 7 girls in a class room. If the names of all those children are written on paper slips and 3 slips are drawn at random from them, then the probability of getting the names of one boy and two girls or one girl and two boys is (A) $\dfrac{1}{5}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Add the two mutually exclusive favourable cases (1 boy+2 girls, 1 girl+2 boys) over the total ways to pick 3 slips from 15; the probability is $4/5$.

   Concept and Intuition

   The two required outcomes ("1 boy & 2 girls" and "1 girl & 2 boys") cannot happen simultaneously in the same draw of 3 slips, so their probabilities simply add (mutually exclusive events).

   Step-by-Step Solution

   1. Total children $=8+7=15$; total ways to draw 3 slips $=\binom{15}{3}=455$.
   2. Ways for 1 boy and 2 girls: $\binom{8}{1}\binom{7}{2}=8\times21=168$.
   3. Ways for 1 girl and 2 boys: $\binom{7}{1}\binom{8}{2}=7\times28=196$.
   4. These are mutually exclusive, so favourable total $=168+196=364$. …

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9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.A basket contains 5 apples and 7 oranges and another basket contains 4 apples and 8 oranges. If one fruit is picked out at random from each basket, then the probability of getting one apple and one orange is (A) $\dfrac{1}{6}$ (B) $\dfrac{7}{18}$ (C) $\dfrac{17}{36}$ (D) $\dfrac{19}{36}$

   ›Reveal solutionSolution

   Sum of two mutually exclusive independent-basket cases gives $\dfrac{17}{36}$, option (C).

   Concept and Intuition

   One fruit is picked from each of two independent baskets. "One apple and one orange" (overall, not specifying which basket gives which) can occur in exactly two distinct, mutually exclusive ways, so their probabilities add.

   Step-by-Step Solution

   1. Basket 1: 5 apples, 7 oranges, total 12. Basket 2: 4 apples, 8 oranges, total 12.
   2. Case (i): apple from basket 1 and orange from basket 2: $P=\dfrac{5}{12}\times\dfrac{8}{12}=\dfrac{40}{144}$.
   3. Case (ii): orange from basket 1 and apple from basket 2: $P=\dfrac{7}{12}\times\dfrac{4}{12}=\dfrac{28}{144}$. …

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10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.For three events $A, B$ and $C$ of a sample space, $P(\text{exactly one of } A \text{ or } B \text{ occurs}) = P(\text{exactly one of } B \text{ or } C \text{ occurs}) = P(\text{exactly one of } C \text{ or } A \text{ occurs}) = \dfrac{1}{4}$. If probability of all the three events occuring simultaneously is $\dfrac{1}{16}$, then the probability that atleast one of the events occur is (A) $\dfrac{3}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{7}{16}$ (D) $\dfrac{7}{32}$

    ›Reveal solutionSolution

    This tests combining the "exactly one of two events" identity with the inclusion-exclusion formula for three events. The probability that at least one event occurs is $\dfrac{7}{16}$.

    Concept and Intuition

    The identity $P(\text{exactly one of }X\text{ or }Y)=P(X)+P(Y)-2P(X\cap Y)$ is the key building block here. Applying it to all three pairs and adding gives a combined relation between $\sum P(\text{single events})$ and $\sum P(\text{pairwise intersections})$ — exactly the two quantities inclusion-exclusion needs (along with the triple intersection) to build $P(A\cup B\cup C)$.

    Step-by-Step Solution

    1. $P(\text{exactly one of }A,B)=P(A)+P(B)-2P(A\cap B)=\tfrac14$.
    2. $P(\text{exactly one of }B,C)=P(B)+P(C)-2P(B\cap C)=\tfrac14$.
    3. $P(\text{exactly one of }C,A)=P(C)+P(A)-2P(C\cap A)=\tfrac14$.
    4. Adding all three: $2[P(A)+P(B)+P(C)]-2[P(AB)+P(BC)+P(CA)]=\tfrac34$.
    5. Divide by 2: $[P(A)+P(B)+P(C)]-[P(AB)+P(BC)+P(CA)]=\tfrac38$. …

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11. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.P, Q and R try to hit the same target one after the other. If their probabilities of hitting the target are $\dfrac{2}{3}, \dfrac{3}{5}, \dfrac{5}{7}$ respectively, then the probability that the target is hit by P or Q but not by R is (A) $\dfrac{26}{105}$ (B) $\dfrac{79}{105}$ (C) 0 (D) $\dfrac{75}{105}$

    ›Reveal solutionSolution

    Independence lets us multiply "(hit by P or Q)" with "(missed by R)".

    Concept and Intuition

    Since the three shooters act independently, the event "target hit by P or Q but not R" factors as $P(\text{P or Q hits})\times P(\text{R misses})$.

    Step-by-Step Solution

    1. $P(\text{P misses})=\tfrac13$, $P(\text{Q misses})=\tfrac25$.
    2. $P(\text{neither P nor Q hits}) = \tfrac13\times\tfrac25=\tfrac{2}{15}$.
    3. $P(\text{P or Q hits}) = 1-\tfrac{2}{15}=\tfrac{13}{15}$.
    4. $P(\text{R misses})=1-\tfrac57=\tfrac27$.
    5. Required probability $=\tfrac{13}{15}\times\tfrac27=\tfrac{26}{105}$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.S is the sample space and A, B are two events of a random experiment. Match the items of List A with the items of List B. List A / List B I. A, B are mutually exclusive events --- a. $P(A \cap B) = P(B) - P(\bar{A})$ II. A, B are independent events --- b. $P(A) \le P(B)$ III. $A \cap B = A$ --- c. $P\left(\dfrac{\bar{A}}{B}\right) = 1 - P(A)$ IV. $A \cup B = S$ --- d. $P(A \cup B) = P(A)+P(B)$ --- e. $P(A)+P(B) = 2$ (A) (I - e) (II - d) (III - c) (IV - b) (B) (I - a) (II - c) (III - e) (IV - b) (C) (I - d) (II - c) (III - b) (IV - a) (D) (I - b) (II - d) (III - a) (IV - c)

    ›Reveal solutionSolution

    Each set/probability condition translates to a probability identity: mutually exclusive $\to$ additive union (d), independence $\to$ conditional-complement rule (c), subset $\to$ probability monotonicity (b), and exhaustive $\to$ the intersection formula (a).

    Concept and Intuition

    Matching set-theoretic conditions on events to their probability consequences is a standard exercise: mutual exclusivity kills the intersection term in $P(A\cup B)$; independence lets conditional probabilities equal unconditional ones; $A\subseteq B$ forces a probability inequality; and $A\cup B=S$ (exhaustive) pins down $P(A\cap B)$ via the inclusion-exclusion formula.

    Step-by-Step Solution

    1. I. Mutually exclusive: $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)-0=P(A)+P(B)$ — this is (d).
    2. II. Independent: $P(A|B)=P(A)$ by definition of independence, so $P(\bar A|B)=1-P(A|B)=1-P(A)$ — this is (c).
    3. III. $A\cap B=A$: this means every outcome in $A$ is also in $B$, i.e. $A\subseteq B$, so $P(A)\le P(B)$ — this is (b). …

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