Q.Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'.

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 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 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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5. 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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6. 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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7. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.A box contains $n$ coins, $m$ of which are fair and the rest are biased. When a biased coin is tossed, the probability of getting a head is twice as likely as tail. A coin is drawn from the box at random and is tossed twice. It is found that first time it shows head and the second time it shows tail. Then the probability that the coin drawn is fair is (A) $\frac{7m}{8n+m}$ (B) $\frac{9m}{8n+m}$ (C) $\frac{7m}{8m+n}$ (D) $\frac{9m}{8m+n}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem; computing the likelihoods for fair vs. biased coins and combining with the prior $m/n$ gives $\frac{9m}{8n+m}$.

   Concept and Intuition

   Bayes' theorem updates our belief about which "type" of coin was drawn, given the observed outcome (head then tail), by weighing each type's prior probability by how likely it was to produce that exact outcome.

   Step-by-Step Solution

   1. Fair coin: $P(H)=P(T)=\frac12$, so $P(HT\mid\text{fair})=\frac12\cdot\frac12=\frac14$.
   2. Biased coin: head is twice as likely as tail, so $P(H)=\frac23,\,P(T)=\frac13$; $P(HT\mid\text{biased})=\frac23\cdot\frac13=\frac29$.
   3. Priors: $P(\text{fair})=\frac mn$, $P(\text{biased})=\frac{n-m}{n}$. …

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8. 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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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 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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12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = Kr^{3}$ where $r = 1, 2, 3, 4$ then ________ (A) $K = \dfrac{1}{100}$ and $P\left(\dfrac{1}{2} < X < \dfrac{5}{2}\,\middle|\,X>1\right) = \dfrac{8}{97}$ (B) $K = \dfrac{1}{99}$ and $P\left(\dfrac{1}{2} < X < \dfrac{5}{2}\,\middle|\,X>1\right) = \dfrac{8}{99}$ (C) $K = \dfrac{1}{100}$ and $P\left(\dfrac{1}{2} < X < \dfrac{5}{2}\,\middle|\,X>1\right) = \dfrac{8}{99}$ (D) $K = \dfrac{1}{100}$ and $P\left(\dfrac{1}{2} < X < \dfrac{5}{2}\,\middle|\,X>1\right) = \dfrac{10}{99}$

    ›Reveal solutionSolution

    First find $K$ from total probability $=1$, then compute a conditional probability by restricting the sample space to $X>1$ and finding which values also satisfy $\tfrac12<X<\tfrac52$.

    Concept and Intuition

    For a discrete probability mass function, probabilities across all values must sum to 1 — this pins down the normalizing constant $K$. A conditional probability $P(E\mid F)$ then equals $P(E\cap F)/P(F)$, restricting attention to outcomes already known to lie in $F$.

    Step-by-Step Solution

    1. $P(X=r)=Kr^3$ for $r=1,2,3,4$. Sum to 1: $K(1+8+27+64)=K(100)=1\Rightarrow K=\dfrac{1}{100}$.
    2. So $P(X=1)=\dfrac{1}{100},\ P(X=2)=\dfrac{8}{100},\ P(X=3)=\dfrac{27}{100},\ P(X=4)=\dfrac{64}{100}$.
    3. Event $F=\{X>1\}=\{2,3,4\}$, with $P(F)=\dfrac{8+27+64}{100}=\dfrac{99}{100}$.
    4. Event $E=\{\tfrac12<X<\tfrac52\}$: since $X$ takes integer values, this means $X\in\{1,2\}$.
    5. $E\cap F = \{2\}$ (since $X=1$ is excluded by $F$). $P(E\cap F)=P(X=2)=\dfrac{8}{100}$. …

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