Q.A black and a red dice are rolled.

Showing the 12 most recent of 69 on this concept.

1. 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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2. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Three boxes $B_1$, $B_2$ and $B_3$ contain balls with different colors as follows:

   	White	Black	Red
   $B_1$	2	1	2
   $B_2$	3	2	4
   $B_3$	4	3	2

   A die is thrown. Box $B_1$ is chosen if either 1 or 2 turns up. Box $B_2$ is chosen if 3 or 4 turns up and box $B_3$ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is drawn at random from that box. If the ball drawn is found to be Red, then the probability that it is drawn from box $B_2$ is (A) $\dfrac{7}{12}$ (B) $\dfrac{5}{12}$ (C) $\dfrac{1}{12}$ (D) $\dfrac{3}{26}$

   ›Reveal solutionSolution

   A Bayes'-theorem problem: reverse the conditional probability "ball is Red, which box?" using the total-probability formula.

   Concept and Intuition

   Bayes' theorem lets us "invert" a conditional probability. We know $P(\text{Red}\mid B_i)$ for each box and the prior $P(B_i)$ (each $1/3$ since the die is fair and each box corresponds to two faces); we want the posterior $P(B_2 \mid \text{Red})$.

   Step-by-Step Solution

   1. Totals in each box: $B_1$: $2+1+2=5$; $B_2$: $3+2+4=9$; $B_3$: $4+3+2=9$.
   2. $P(\text{Red}\mid B_1)=\dfrac{2}{5}$, $P(\text{Red}\mid B_2)=\dfrac{4}{9}$, $P(\text{Red}\mid B_3)=\dfrac{2}{9}$.
   3. Each box has prior probability $\dfrac13$ (each box corresponds to two die faces out of six).
   4. Total probability: $P(\text{Red}) = \dfrac13\left(\dfrac25+\dfrac49+\dfrac29\right) = \dfrac13\left(\dfrac25+\dfrac23\right) = \dfrac13\cdot\dfrac{16}{15}=\dfrac{16}{45}$. …

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3. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.A pair of dice is thrown. Then the probability that either of the dice shows 2 when their sum is 6 is (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   This is a conditional probability: restrict the sample space to pairs summing to 6, then count how many of those have a 2 on either die.

   Concept and Intuition

   "Given that the sum is 6" restricts attention to only those (die1, die2) pairs whose total is 6. Within that restricted, equally-likely set, we simply count the favourable outcomes.

   Step-by-Step Solution

   1. Pairs $(d_1,d_2)$ with $d_1+d_2=6$: $(1,5),(2,4),(3,3),(4,2),(5,1)$ — 5 outcomes, each equally likely given the sum is 6.
   2. Favourable outcomes (either die shows 2): $(2,4)$ and $(4,2)$ — 2 outcomes.
   3. Required probability $=\dfrac{2}{5}$.

   Common Mistakes …

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4. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.Two persons A and B throw three unbiased dice one after the another. If A gets the sum 13, then the probability that B gets higher sum is (A) $\dfrac{5}{216}$ (B) $\dfrac{4}{27}$ (C) $\dfrac{35}{216}$ (D) $\dfrac{20}{216}$

   ›Reveal solutionSolution

   This tests knowing (or deriving) the distribution of the sum of three dice and just needs $P(\text{sum}>13)$ since B's throw is independent of A's. Answer: $35/216$.

   Concept and Intuition

   Since the two people throw independently, "A got 13" tells us nothing about B's throw — we just need $P(\text{sum of 3 dice}>13)=P(\text{sum}\ge14)$ out of all $6^3=216$ equally likely outcomes.

   Step-by-Step Solution

   1. Total outcomes for three dice: $6^3=216$.
   2. The number of ways to get sum $s$ with 3 dice is symmetric: $N(s)=N(21-s)$ (since replacing each die value $v$ by $7-v$ maps sum $s$ to $21-s$).
   3. Known counts: $N(3)=1,N(4)=3,N(5)=6,N(6)=10,N(7)=15$; by symmetry $N(18)=1,N(17)=3,N(16)=6,N(15)=10,N(14)=15$. …

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5. 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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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability that both are odd is (A) $\dfrac{3}{8}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{8}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Conditioning on 'sum even' restricts to same-parity pairs; among those, the odd-odd pairs make up $5/8$.

   Concept and Intuition

   Two digits sum to an even number exactly when they have the same parity (both odd or both even) — an odd plus an even is always odd. This turns the conditional-probability question into a simple counting problem over same-parity pairs only.

   Step-by-Step Solution

   1. Digits $1$ through $9$: odd $=\{1,3,5,7,9\}$ (5 digits), even $=\{2,4,6,8\}$ (4 digits).
   2. Sum is even $\iff$ both digits are odd, or both are even.
   3. Number of ways to choose 2 odd digits: $\binom52=10$.
   4. Number of ways to choose 2 even digits: $\binom42=6$.
   5. Total ways with even sum: $10+6=16$. …

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7. 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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8. 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). …

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9. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.One ticket is selected at random from 50 tickets numbered $00, 01, 02, \ldots 49$. The probability that sum of the digits is 10, given that product of the digits is 9 is (A) $\dfrac{9}{10}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{25}$

   ›Reveal solutionSolution

   Only two tickets (19 and 33) have digit-product 9, and only one of them (19) also has digit-sum 10, giving conditional probability $1/2$.

   Concept and Intuition

   Conditional probability $P(\text{sum}=10\mid\text{product}=9)$ restricts attention entirely to the tickets satisfying the "given" condition (product $=9$), then asks what fraction of those also satisfy the target condition (sum $=10$).

   Step-by-Step Solution

   1. List all two-digit combinations $(d_1,d_2)$ with $d_1\in\{0,1,2,3,4\}$ (tens digit, since tickets run 00–49) and $d_2\in\{0,\dots,9\}$ (units digit) whose product is 9:
      • $d_1=1,d_2=9$: ticket $19$.
      • $d_1=3,d_2=3$: ticket $33$.
      • No other integer pairs with $d_1\le4$ give product 9.
   2. So the "product = 9" event has exactly 2 tickets: $\{19, 33\}$.
   3. Check digit sums: $19\to1+9=10$ ✓; $33\to3+3=6$ ✗.
   4. Only 1 out of these 2 tickets also has digit-sum 10. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.A person is known to speak the truth in 3 out of 4 occasions. If he throws a die and reports that it is six, then the probability that it is actually six is (A) $\dfrac{3}{8}$ (B) $\dfrac{2}{7}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{4}{5}$

    ›Reveal solutionSolution

    A classic Bayes'-theorem problem: combine the prior $P(\text{six})=1/6$ with the witness's reliability $3/4$ to get the posterior probability it's actually six, given he reports six. The answer is $3/8$.

    Concept and Intuition

    Before he speaks, the die has $P(\text{six})=1/6$. His report is evidence, but he isn't perfectly reliable (truthful $3/4$ of the time). Bayes' theorem updates the prior probability using how likely the report "six" is under each of the two possibilities (six actually occurred vs. it didn't).

    Step-by-Step Solution

    1. Let $E_1$ = the die actually shows six, $E_2$ = it doesn't. $P(E_1)=\dfrac16$, $P(E_2)=\dfrac56$.
    2. Let $A$ = he reports "six". If $E_1$ occurred, he reports six truthfully with probability $\dfrac34$: $P(A\mid E_1)=\dfrac34$.
    3. If $E_2$ occurred, he reports "six" only if he lies, with probability $\dfrac14$: $P(A\mid E_2)=\dfrac14$.
    4. By Bayes' theorem:

    $P(E_1\mid A)=\frac{P(A\mid E_1)P(E_1)}{P(A\mid E_1)P(E_1)+P(A\mid E_2)P(E_2)}.$

    5. Numerator: $\dfrac34\cdot\dfrac16=\dfrac{3}{24}=\dfrac18$.
    6. Denominator's second term: $\dfrac14\cdot\dfrac56=\dfrac{5}{24}$.
    7. Sum $=\dfrac{3}{24}+\dfrac{5}{24}=\dfrac{8}{24}=\dfrac13$. …

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11. 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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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.A bag contains 2 white, 3 green and 5 red balls. If three balls are drawn one after the other without replacement, then the probability that the last ball drawn was red is (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{5}{9}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    By the symmetry of random sampling without replacement, the probability that the last ball drawn is red equals the overall fraction of red balls in the bag: $\dfrac{5}{10}=\dfrac12$.

    Concept and Intuition

    When balls are drawn one after another without replacement, every ball is equally likely to occupy any given position in the drawing order (all $10!$ orderings of the bag's balls are equally likely). Hence the marginal probability that the ball in any fixed position (first, second, ..., last) is red is just (number of red balls)/(total balls), independent of which position we pick.

    Step-by-Step Solution

    1. Bag: $2$ white $+3$ green $+5$ red $=10$ balls total.
    2. Consider a full random permutation of all 10 balls (drawing three is just looking at the first three positions of such a permutation, but the argument works for any position).
    3. By symmetry, $P(\text{ball in position }k\text{ is red})=\dfrac{5}{10}$ for every position $k$, including the third (last) position drawn here. …

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