Q.Determine $P(E|F)$. A die is thrown three times, $E$ : 4 appears on the third toss, $F$ : 6 and 5 appears respectively on first two tosses.

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 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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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 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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5. 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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6. 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. …

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7. 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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8. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.An unbiased coin is tossed 3 times. If the third toss gets head, then the probability of getting at least one more head is (A) $3/4$ (B) $1/4$ (C) $1/2$ (D) $1/3$

   ›Reveal solutionSolution

   This tests recognizing conditional independence: given the third toss's outcome, the first two tosses remain independent unbiased flips, so their probability is unaffected by the condition. Answer: $3/4$.

   Concept and Intuition

   Since coin tosses are independent events, knowing the outcome of the third toss (head) gives us no information about the first two tosses. So the question really just asks: what's the probability of getting at least one head in two independent fair coin tosses? This is most easily computed via the complement (no heads at all in two tosses).

   Step-by-Step Solution

   1. The condition 'the third toss gets head' is independent of the outcomes of the first two tosses (each coin toss is independent of the others).
   2. So conditioning on the third toss being heads doesn't change the probabilities for the first two tosses — they remain two independent fair coin flips.
   3. We want $P(\text{at least one head among the first two tosses})$.
   4. Use the complement: $P(\text{no heads in first two tosses}) = P(\text{both tails}) = \dfrac12\times\dfrac12=\dfrac14$.
   5. So $P(\text{at least one head}) = 1-\dfrac14 = \dfrac34$.

   Common Mistakes …

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9. 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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10. 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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11. 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. $P(F_2\cap\text{girl})=\dfrac13\cdot\dfrac23=\dfrac29$. …

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12. 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: $k=3:\binom33=1\Rightarrow1/20$; $k=4:\binom43=4\Rightarrow4/20$; $k=5:\binom53=10\Rightarrow10/20$; $k=6:\binom63=20\Rightarrow20/20=1$. …

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