Q.If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then (A) $P(B|A) = 1$ (B) $P(A|B) = 1$ (C) $P(B|A) = 0$ (D) $P(A|B) = 0$

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A die is thrown three times. If the sum of the numbers thrown is 15, then the probability that the first throw was a Four, is (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{5}{108}$ (D) $\dfrac{1}{108}$

   ›Reveal solutionSolution

   A conditional probability found by directly counting favourable die-triples against all triples summing to the given total.

   Concept and Intuition

   $P(\text{first}=4\mid\text{sum}=15)=\dfrac{\#\{\text{triples with first}=4,\ \text{sum}=15\}}{\#\{\text{triples with sum}=15\}}$ — a straightforward application of conditional probability by counting, since all $6^3$ triples are equally likely.

   Step-by-Step Solution

   1. If the first throw is $4$, the other two throws (each from $1$ to $6$) must sum to $15-4=11$.
   2. Pairs of dice summing to $11$: $(5,6)$ and $(6,5)$ — 2 ways.
   3. Now count all triples $(a,b,c)$, each in $1$–$6$, with $a+b+c=15$. Substitute $a'=6-a,\,b'=6-b,\,c'=6-c$ (each in $0$–$5$): then $a'+b'+c'=18-15=3$.
   4. Number of non-negative integer solutions to $a'+b'+c'=3$ is $\binom{3+2}{2}=10$; since $3<5$ none violate the upper bound of $5$, so all $10$ are valid.
   5. So there are $10$ triples with sum $15$ in total. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In a college, the heights of 4% of male and 1% of female students are more than 1.8 meters. If 60% of the total students are female and a student selected at random has height more than 1.8 meters, then the probability that this student is a female, is (A) $\dfrac{8}{11}$ (B) $\dfrac{6}{11}$ (C) $\dfrac{5}{11}$ (D) $\dfrac{3}{11}$

   ›Reveal solutionSolution

   A direct application of Bayes' theorem: given the height is above 1.8m, find the probability the student is female.

   Concept and Intuition

   This is the classic "reverse conditional probability" setup — we know $P(\text{tall}\mid\text{gender})$ for each gender and the gender proportions, and want $P(\text{gender}\mid\text{tall})$. Bayes' theorem converts one direction of conditioning into the other by weighting each gender's tall-probability by its population share and normalizing.

   Step-by-Step Solution

   1. Let $M$ = male, $F$ = female, $H$ = height $>1.8$m. Given: $P(F)=0.6$, $P(M)=0.4$, $P(H\mid M)=0.04$, $P(H\mid F)=0.01$.
   2. Total probability of height $>1.8$m: $P(H)=P(M)P(H\mid M)+P(F)P(H\mid F)=0.4(0.04)+0.6(0.01)=0.016+0.006=0.022$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A bag 'A' contains 2 black and 3 white balls. Another bag 'B' contains 3 black and 2 white balls. Two balls are drawn randomly from 'A' and placed in 'B'. Later, if two balls are drawn randomly from 'B', then the probability of getting a black ball and a white ball from it is (A) $\dfrac{59}{105}$ (B) $\dfrac{46}{105}$ (C) $\dfrac{59}{210}$ (D) $\dfrac{67}{210}$

   ›Reveal solutionSolution

   A two-stage random-transfer problem — condition on what got transferred from A to B, then apply the law of total probability.

   Concept and Intuition

   Since the composition of bag B after the transfer depends on which 2 balls were drawn from A, we must split into the three possible transfer outcomes (BB, BW, WW), compute the probability of each, then the conditional probability of drawing one black and one white ball from the resulting bag B, and combine via the law of total probability.

   Step-by-Step Solution

   1. Bag A has 5 balls (2B, 3W); ways to choose 2: $\binom{5}{2}=10$.
      • $P(\text{BB from A}) = \binom{2}{2}/\binom{5}{2} = 1/10$.
      • $P(\text{BW from A}) = \binom{2}{1}\binom{3}{1}/\binom{5}{2} = 6/10 = 3/5$.
      • $P(\text{WW from A}) = \binom{3}{2}/\binom{5}{2} = 3/10$.
   2. Bag B originally has 3B, 2W (5 balls); after adding 2 balls it has 7 balls, and we want $P(1B,1W)$ drawn from it: $\binom{7}{2}=21$ ways.
      • If BB added: B = 5B, 2W. $P(1B,1W) = \binom{5}{1}\binom{2}{1}/21 = 10/21$.
      • If BW added: B = 4B, 3W. $P(1B,1W) = \binom{4}{1}\binom{3}{1}/21 = 12/21$.
      • If WW added: B = 3B, 4W. $P(1B,1W) = \binom{3}{1}\binom{4}{1}/21 = 12/21$.
   3. Total probability: …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If two cards are drawn at a time at random from a well shuffled pack of 52 playing cards and they are found to be a king card and a card with prime number, then the probability that they are a black king card and a card with an odd prime number is (A) $\dfrac{32}{663}$ (B) $\dfrac{12}{663}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{5}{8}$

   ›Reveal solutionSolution

   A conditional-probability counting problem: restrict the sample space to (King, prime-numbered-card) pairs, then count how many of those pairs are (black King, odd-prime card).

   Concept and Intuition

   A standard deck has 4 suits (2 black: spades, clubs; 2 red: hearts, diamonds), each with 13 ranks: A, 2–10, J, Q, K. "Cards with a prime number" means cards whose rank value is a prime, i.e. rank 2, 3, 5, or 7 — four ranks × 4 suits = 16 cards. Kings are a separate rank (not a "number" card), 4 total, 2 of them black (spade, club). Since we are told the two drawn cards are exactly one King and one prime-numbered card, we treat every (King, prime-card) pairing as equally likely and count favourable outcomes among them.

   Step-by-Step Solution

   1. Total King cards = 4; total prime-numbered cards (ranks 2,3,5,7) = $4\times4=16$.
   2. Sample space size (ways to have one King and one prime card) = $4\times16=64$.
   3. "Odd prime number" cards are ranks 3, 5, 7 (2 is the only even prime, excluded): $3\times4=12$ cards.
   4. Black Kings = 2 (spade King, club King). …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Bag A contains 3 white and 4 black balls. Bag B contains 4 white and 3 black balls. Bag C contains 2 white and 5 black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag C is (A) $\dfrac{1}{6}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{13}$

   ›Reveal solutionSolution

   This is a direct Bayes'-theorem (inverse probability) question. Answer: $\dfrac{2}{9}$.

   Concept and Intuition

   We're given the outcome (a white ball was drawn) and asked for the probability of a particular cause (it came from bag C). This is exactly Bayes' theorem: $P(C\mid W) = \dfrac{P(C)P(W\mid C)}{\sum_i P(\text{bag}_i)P(W\mid \text{bag}_i)}$.

   Step-by-Step Solution

   1. Each bag is chosen with probability $\frac13$.
   2. $P(W\mid A) = \dfrac{3}{7}$ (3 white out of 7 total in bag A).
   3. $P(W\mid B) = \dfrac{4}{7}$ (4 white out of 7 in bag B).
   4. $P(W\mid C) = \dfrac{2}{7}$ (2 white out of 7 in bag C).
   5. Total probability of white: $P(W) = \frac13\left(\dfrac37+\dfrac47+\dfrac27\right) = \frac13\cdot\dfrac{9}{7} = \dfrac{9}{21} = \dfrac{3}{7}$. …

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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In a bolt factory, machines A, B, C manufacture 25%, 35%, 40% of the total output respectively. There is a chance of having 5%, 4%, 2% defective bolts manufactured by A, B, C respectively. If a bolt is drawn at random from the output, then the probability that it is defective is (A) $\dfrac{69}{2000}$ (B) $\dfrac{59}{2000}$ (C) $\dfrac{79}{2000}$ (D) $\dfrac{89}{2000}$

   ›Reveal solutionSolution

   Classic total-probability ("law of total probability") setup: weight each machine's defect rate by its share of output and sum. The answer is (A).

   Concept and Intuition

   When an item can come from several mutually exclusive, exhaustive sources (here, machines A, B, C), and each source has its own conditional probability of producing a defect, the overall probability of a defect is the weighted average:

   $P(D) = \sum_i P(\text{source}_i)\,P(D\mid \text{source}_i).$

   This is the law of total probability — it's the natural way to combine "how much each machine contributes" with "how likely each machine's own output is defective."

   Step-by-Step Solution

   1. Let $A, B, C$ denote drawing a bolt from each machine: $P(A)=0.25$, $P(B)=0.35$, $P(C)=0.40$ (these sum to 1, as they must).
   2. Conditional defect probabilities: $P(D\mid A)=0.05$, $P(D\mid B)=0.04$, $P(D\mid C)=0.02$.
   3. By the law of total probability:

   $P(D) = P(A)P(D\mid A) + P(B)P(D\mid B) + P(C)P(D\mid C).$

   4. Compute each term: $0.25\times0.05=0.0125$; $0.35\times0.04=0.014$; $0.40\times0.02=0.008$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If it is known that a woman has two children and she has at least one girl child, the probability that the woman has both girl children is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   This is a classic conditional-probability trap: conditioning on "at least one girl" (not "the elder/first child is a girl") leaves 3 equally likely outcomes, of which 1 is both-girls, giving $\dfrac13$.

   Concept and Intuition

   With two children, birth order matters for counting equally likely outcomes: $BB, BG, GB, GG$, each with probability $1/4$. The event "at least one girl" is a set of outcomes, not a statement about a specific (e.g. first) child, so it correctly removes only $BB$ and keeps three outcomes, not two. This distinguishes it from the simpler (and different) question "given the elder child is a girl," which would leave only $\{GB,GG\}$ and give probability $1/2$.

   Step-by-Step Solution

   1. Sample space (ordered by birth, say elder-younger): $\{BB,BG,GB,GG\}$, each with probability $\dfrac14$.
   2. Event $E$ = "at least one girl" = $\{BG,GB,GG\}$, so $P(E)=\dfrac34$. …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A bag P contains 5 white and 4 blue balls. Another bag Q contains 4 white and 5 blue balls. A bag is randomly selected and a ball is drawn at random. If the ball selected from that bag is transferred to another bag, then the probability that bag Q has same number of blue and white balls is (A) $2/9$ (B) $1/9$ (C) $5/9$ (D) $4/9$

   ›Reveal solutionSolution

   This tests conditional probability with a transfer-between-bags setup; you must track which single transfer restores balance to bag Q, in both directions of transfer. Answer: $5/9$.

   Concept and Intuition

   A ball moves from whichever bag is picked into the other bag. "Bag Q ends up balanced" can happen two structurally different ways: either Q gains a ball (if P was picked) or Q loses a ball (if Q was picked). Each scenario needs the transferred ball to be a specific colour for Q's counts to equalize, so we compute each scenario's probability and add them (mutually exclusive cases, law of total probability).

   Step-by-Step Solution

   1. Bag P: $5W,4B$ (9 total). Bag Q: $4W,5B$ (9 total).
   2. Scenario A (P picked, prob $1/2$): a ball moves P→Q, so Q becomes a 10-ball bag. Q needs to end at $5W,5B$. Since Q started $4W,5B$, the incoming ball must be white. $P(\text{white}\mid P)=5/9$. Probability of this scenario succeeding: $\frac12\times\frac59=\frac{5}{18}$. …

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9. 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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10. 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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11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it randomly. If the number on the card is found to be a non-prime number, the probability that the card was drawn from Box I is (A) $\dfrac{4}{17}$ (B) $\dfrac{8}{17}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{2}{3}$

    ›Reveal solutionSolution

    Bayes' theorem with the counts of non-prime numbers in each box gives $P(\text{Box I}\mid\text{non-prime})=8/17$.

    Concept and Intuition

    This is a classic "which urn/box did it come from" Bayes' problem: we're given the outcome (a non-prime card) and asked to find the probability of the cause (which box). We need $P(\text{non-prime}\mid\text{box})$ for each box, weighted by the prior $P(\text{box})=1/2$.

    Step-by-Step Solution

    1. Primes from 1 to 30: $2,3,5,7,11,13,17,19,23,29$ — that's $10$ primes, so $30-10=20$ non-primes in Box I.
    2. Primes from 31 to 50: $31,37,41,43,47$ — that's $5$ primes, so $20-5=15$ non-primes in Box II.
    3. $P(\text{non-prime}\mid\text{Box I})=\dfrac{20}{30}=\dfrac23$, $P(\text{non-prime}\mid\text{Box II})=\dfrac{15}{20}=\dfrac34$.
    4. $P(\text{Box I})=P(\text{Box II})=\dfrac12$.
    5. $P(\text{Box I and non-prime})=\dfrac12\cdot\dfrac23=\dfrac13$; $P(\text{Box II and non-prime})=\dfrac12\cdot\dfrac34=\dfrac38$. …

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12. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.Two balls are drawn at random from a box containing 4 white, 6 black balls one after the other without replacement. If it is known that second ball drawn is black, then the probability that the first ball drawn is also black is (A) $\dfrac{5}{11}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{5}{13}$

    ›Reveal solutionSolution

    Conditional probability with sampling without replacement, solved with Bayes' theorem; answer is $\frac{5}{9}$.

    Concept and Intuition

    When balls are drawn one after another without replacement, the marginal probability that any particular draw (say the 2nd) is black equals the overall proportion of black balls, $\frac{6}{10}$ — position doesn't matter for the marginal event by symmetry. To find the conditional probability of the first draw given information about the second, we use Bayes' theorem: we need the joint probability of both events and divide by the marginal probability of the conditioning event.

    Step-by-Step Solution

    1. Total balls: 4 white (W) + 6 black (B) = 10.
    2. $P(\text{2nd is black}) = P(\text{1st B, 2nd B}) + P(\text{1st W, 2nd B})$ $= \frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{6}{9} = \frac{30}{90}+\frac{24}{90} = \frac{54}{90} = \frac{3}{5}$.
    3. $P(\text{1st black and 2nd black}) = \frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90} = \frac{1}{3}$. …

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