Q.(a) A die with numbers 1 to 6 is biased such that $P(2) = \dfrac{3}{10}$ and the probability of other numbers is equal. Find the mean of the number of times the number 2 appears on the die, if the die is thrown twice.

Showing the 12 most recent of 73 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In a binomial distribution $(q+p)^n$, if $q - p = \dfrac{1}{4}$ and the product of its mean and the variance is $\dfrac{45}{8}$, then $P(X=2) =$ (A) $\dfrac{(126)5^6}{8^8}$ (B) $\dfrac{(252)5^6}{8^8}$ (C) $\dfrac{(63)3^5}{5^6}$ (D) $\dfrac{(72)3^5}{5^6}$

   ›Reveal solutionSolution

   Use $q-p$ and $p+q=1$ to find $p,q$, then the given mean$\times$variance product to find $n$, and finally compute $P(X=2)$ from the binomial formula.

   Concept and Intuition

   For a binomial distribution with parameters $n,p$ (and $q=1-p$): mean $=np$ and variance $=npq$. Their product $n^2p^2q$ is a clean single expression in $n$ once $p,q$ are known numerically, letting us solve for $n$ without needing to separately isolate mean and variance.

   Step-by-Step Solution

   1. From $p+q=1$ and $q-p=\dfrac14$: adding, $2q=\dfrac54\Rightarrow q=\dfrac58$; then $p=1-\dfrac58=\dfrac38$.
   2. Mean $=np$, Variance $=npq$. Their product: $(np)(npq)=n^2p^2q$.
   3. $p^2q=\left(\dfrac38\right)^2\cdot\dfrac58=\dfrac{9}{64}\cdot\dfrac58=\dfrac{45}{512}$.
   4. Given $n^2p^2q=\dfrac{45}{8}$: $n^2=\dfrac{45/8}{45/512}=\dfrac{512}{8}=64\Rightarrow n=8$.
   5. $P(X=2)=\binom{8}{2}p^2q^{6}=28\left(\dfrac38\right)^2\left(\dfrac58\right)^6$.
   6. $\left(\dfrac38\right)^2=\dfrac9{64}$, $\left(\dfrac58\right)^6=\dfrac{5^6}{8^6}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $X \sim B(n,p)$ is a binomial variate, $8p^2 + 15p - 2 = 0$ and the product of the mean and variance of $X$ is $\dfrac{7}{8}$, then $P(X=4) =$ (A) ${}^6C_4 \dfrac{5^2}{6^6}$ (B) ${}^7C_4 \dfrac{4^3}{5^7}$ (C) ${}^5C_4 \dfrac{3}{4^5}$ (D) ${}^8C_4 \dfrac{7^4}{8^8}$

   ›Reveal solutionSolution

   Solve the quadratic for $p$, use $(\text{mean})(\text{variance}) = n^2p^2q$ to find $n$, then apply the binomial pmf. Answer: ${}^8C_4\dfrac{7^4}{8^8}$.

   Concept and Intuition

   For $X\sim B(n,p)$: mean $=np$, variance $=npq$ where $q=1-p$. Their product is $n^2p^2q$, a single equation in $n$ once $p$ is known from the quadratic.

   Step-by-Step Solution

   1. Solve $8p^2+15p-2=0$: $p = \dfrac{-15\pm\sqrt{225+64}}{16} = \dfrac{-15\pm17}{16}$. The valid probability root is $p=\dfrac{2}{16}=\dfrac18$ (the other root is negative).
   2. So $q = 1-p = \dfrac78$.
   3. Mean $\times$ Variance $= (np)(npq) = n^2p^2q = n^2\cdot\dfrac{1}{64}\cdot\dfrac78 = \dfrac{7n^2}{512}$.
   4. Set equal to $\dfrac78$: $\dfrac{7n^2}{512} = \dfrac78 \Rightarrow n^2 = \dfrac{512}{8} = 64 \Rightarrow n = 8$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Let $E_1, E_2$ and $E_3$ be mutually independent events Statement I: $E_1$ and $E_2 \cup E_3$ are independent Statement II: $E_1$ and $E_2 \cap E_3$ are independent Which one of the following options is correct? (A) Both I and II are true (B) Only I is true (C) Only II is true (D) Both I and II are false

   ›Reveal solutionSolution

   Mutual independence of $E_1,E_2,E_3$ guarantees $E_1$ is independent of every Boolean combination of $E_2,E_3$ (unions, intersections, complements). Both statements are true — option (A).

   Concept and Intuition

   Mutual independence of three events means every subset of them satisfies the product rule: $P(E_i\cap E_j)=P(E_i)P(E_j)$ for all pairs, and $P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2)P(E_3)$. A standard consequence is that any event generated from $E_2$ and $E_3$ alone (via $\cup,\cap,{}^c$) remains independent of $E_1$, because independence of $E_1$ from $E_2$ and from $E_3$ combines nicely through inclusion-exclusion and De Morgan's laws.

   Step-by-Step Solution

   1. Statement I: Check $P(E_1\cap(E_2\cup E_3)) = P(E_1)P(E_2\cup E_3)$.

   $P(E_1\cap(E_2\cup E_3)) = P((E_1\cap E_2)\cup(E_1\cap E_3)) = P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3)$

   $= P(E_1)P(E_2)+P(E_1)P(E_3)-P(E_1)P(E_2)P(E_3) \quad(\text{using mutual independence})$

   $= P(E_1)\left[P(E_2)+P(E_3)-P(E_2)P(E_3)\right] = P(E_1)\,P(E_2\cup E_3).$

   So Statement I is true.

   2. Statement II: Check $P(E_1\cap(E_2\cap E_3)) = P(E_1)P(E_2\cap E_3)$.

   $P(E_1\cap E_2\cap E_3) = P(E_1)P(E_2)P(E_3) \quad(\text{mutual independence, directly})$ …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A pair of dice is thrown independently 3 times. The probability of getting a total score of at least 9 twice, is (A) $\dfrac{925}{5832}$ (B) $\dfrac{975}{5832}$ (C) $\dfrac{1025}{5832}$ (D) $\dfrac{1075}{5832}$

   ›Reveal solutionSolution

   First find the single-throw probability of a total $\ge9$ with a pair of dice, then apply the binomial probability formula for exactly 2 successes in 3 independent trials. The answer is (B).

   Concept and Intuition

   Each throw of a pair of dice is an independent trial with a fixed "success" probability $p=P(\text{sum}\ge9)$. Asking for the probability of exactly 2 successes in 3 independent trials is a textbook binomial distribution question:

   $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}.$

   Step-by-Step Solution

   1. Count outcomes of two dice giving sum $\ge9$ (out of 36 total equally likely outcomes):
      • sum $=9$: $(3,6),(4,5),(5,4),(6,3)$ → 4 ways
      • sum $=10$: $(4,6),(5,5),(6,4)$ → 3 ways
      • sum $=11$: $(5,6),(6,5)$ → 2 ways
      • sum $=12$: $(6,6)$ → 1 way
      • Total favorable $=4+3+2+1=10$.
   2. Single-throw probability: $p=\dfrac{10}{36}=\dfrac{5}{18}$, and $q=1-p=\dfrac{13}{18}$.
   3. We want exactly 2 successes in $n=3$ independent throws: $P(X=2)=\binom{3}{2}p^2q^1 = 3\cdot\left(\frac{5}{18}\right)^2\cdot\frac{13}{18}.$ …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $A, B, C$ are independent events of a random experiment such that $P(A) = 3/4$, $P(B) = 5/6$ and $P(C) = 2/3$, then the probability that exactly one of the events occur is (A) $\dfrac{1}{4}$ (B) $\dfrac{5}{36}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{71}{72}$

   ›Reveal solutionSolution

   Sum the three mutually exclusive ways exactly one of three independent events can occur. Answer: $\dfrac5{36}$.

   Concept and Intuition

   For independent events, "exactly one occurs" means one event happens and the other two fail, and there are three such disjoint scenarios (which one occurs). Because the events are independent, each scenario's probability is simply the product of the individual (or complementary) probabilities, and we add the three scenarios since they can't happen simultaneously.

   Step-by-Step Solution

   1. Complements: $P(A')=1-\frac34=\frac14$, $P(B')=1-\frac56=\frac16$, $P(C')=1-\frac23=\frac13$.
   2. Scenario "only $A$": $P(A)P(B')P(C')=\frac34\cdot\frac16\cdot\frac13=\frac{3}{72}$.
   3. Scenario "only $B$": $P(A')P(B)P(C')=\frac14\cdot\frac56\cdot\frac13=\frac{5}{72}$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A, B, C are three independent events of a random experiment such that $P(C) = 2/3$. If probabilities of occurrence of only A, only B and only C are respectively $\dfrac{4}{60}, \dfrac{3}{60}$ and $\dfrac{2}{60}$, then the ratio of the probability of the occurrence of A to the occurrence of B is (A) $16:15$ (B) $4:3$ (C) $12:25$ (D) $2:3$

   ›Reveal solutionSolution

   This tests setting up "only-one-event" probabilities for independent events as products with complements, then solving the resulting system for $P(A)$ and $P(B)$.

   Concept and Intuition

   For independent events, "only $A$ occurs" means $A$ happens and both $B,C$ fail: $P(A)(1-P(B))(1-P(C))$. Writing this out for all three "only" events gives three equations in the two unknowns $p=P(A)$ and $q=P(B)$ (since $P(C)$ is already known) — enough to solve completely, with the third equation serving as a consistency check.

   Step-by-Step Solution

   1. Let $p=P(A)$, $q=P(B)$, and $P(C)=\frac23$ so $1-P(C)=\frac13$.
   2. Only $A$: $p(1-q)\cdot\frac13=\frac4{60}=\frac1{15}\Rightarrow p(1-q)=\frac15$.
   3. Only $B$: $(1-p)q\cdot\frac13=\frac3{60}=\frac1{20}\Rightarrow(1-p)q=\frac3{20}$.
   4. Only $C$: $(1-p)(1-q)\cdot\frac23=\frac2{60}=\frac1{30}\Rightarrow(1-p)(1-q)=\frac1{20}$.
   5. From steps 2 and 3: $p-pq=\frac15$ and $q-pq=\frac3{20}$; subtracting, $p-q=\frac15-\frac3{20}=\frac1{20}$, so $p=q+\frac1{20}$.
   6. Expand step 4: $1-p-q+pq=\frac1{20}$. Using $pq=p-\frac15$ from step 2: $1-p-q+p-\frac15=\frac1{20}\Rightarrow\frac45-q=\frac1{20}\Rightarrow q=\frac45-\frac1{20}=\frac{15}{20}=\frac34$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $X \sim B\left(n, \dfrac{1}{4}\right)$, $P(X=2) = P(X=3)$ and $\sum_{K=0}^{2} P(X=K) = \dfrac{3^9}{4^{11}} M$, then $M =$ (A) 97 (B) 38 (C) 128 (D) 152

   ›Reveal solutionSolution

   This tests binomial-term ratios to pin down $n$, then a direct sum of the first three binomial terms expressed as a common power of $3/4$. Answer: $M=97$.

   Concept and Intuition

   For $X\sim B(n,p)$, consecutive term ratios satisfy $\dfrac{P(X=k+1)}{P(X=k)}=\dfrac{n-k}{k+1}\cdot\dfrac{p}{1-p}$. Setting the given equal terms lets us solve for $n$ directly. Once $n$ is fixed, each of $P(X=0),P(X=1),P(X=2)$ can be written as a multiple of the same power $(3/4)^9$, so their sum collapses into one clean coefficient times that common factor — matching the given closed form.

   Step-by-Step Solution

   1. $P(X=2)=P(X=3)$ with $p=1/4$: $\dfrac{n-2}{3}\cdot\dfrac{1/4}{3/4}=1 \Rightarrow \dfrac{n-2}{9}=1 \Rightarrow n=11$.
   2. With $n=11$: $P(X=0)=(3/4)^{11}$, $P(X=1)=11\cdot\frac14\cdot(3/4)^{10}$, $P(X=2)=\binom{11}{2}\cdot\frac{1}{16}\cdot(3/4)^9$.
   3. Factor out $(3/4)^9$: $(3/4)^{11}=(3/4)^9\cdot\frac{9}{16}$; $(3/4)^{10}=(3/4)^9\cdot\frac34$, so $P(X=1)=(3/4)^9\cdot 11\cdot\frac14\cdot\frac34 = (3/4)^9\cdot\frac{33}{16}$; and $P(X=2)=(3/4)^9\cdot 55\cdot\frac{1}{16}=(3/4)^9\cdot\frac{55}{16}$. …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A hunter is firing at a target. He has only 10% chance of firing (hitting) it in one round. The number of rounds he must fire in order to have at least 50% chance of hitting the target at least once, is (A) $8$ (B) $7$ (C) $6$ (D) $5$

   ›Reveal solutionSolution

   Use the complement (probability of missing every round) to find the smallest $n$ with $(0.9)^n\le0.5$.

   Concept and Intuition

   "At least one hit" is most easily handled via its complement, "no hits at all" — the rounds are independent, so the probability of missing all $n$ rounds is $(0.9)^n$, and we want this to drop to at most $0.5$.

   Step-by-Step Solution

   1. $P(\text{hit in one round})=0.1\Rightarrow P(\text{miss})=0.9$.
   2. $P(\text{at least one hit in }n\text{ rounds})=1-(0.9)^n$.
   3. Require $1-(0.9)^n\ge0.5\Rightarrow(0.9)^n\le0.5$.
   4. Compute powers: $0.9^5=0.59049$, $0.9^6=0.531441$, $0.9^7=0.4782969$. …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.In a binomial distribution consisting of 5 independent trials, the probabilities of 1 success and 2 successes are 0.4096 and 0.2048 respectively. Then the variance of this distribution is (A) $0.80$ (B) $0.75$ (C) $0.64$ (D) $0.72$

   ›Reveal solutionSolution

   Using the ratio of two consecutive binomial probabilities eliminates the binomial constant and gives $p/q$ directly; the variance is then $npq = 0.80$.

   Concept and Intuition

   In a Binomial$(n,p)$ distribution, $P(X=k) = \binom{n}{k}p^k q^{n-k}$ where $q=1-p$. Taking the ratio of consecutive terms $P(k+1)/P(k)$ cancels most of the structure and leaves a simple expression in $p,q$ — a standard trick to pin down $p$ without solving messy equations for each probability separately.

   Step-by-Step Solution

   1. Here $n=5$. $P(X=1) = \binom{5}{1}pq^4 = 5pq^4 = 0.4096$ and $P(X=2) = \binom{5}{2}p^2q^3 = 10p^2q^3 = 0.2048$.
   2. Form the ratio: $\dfrac{P(X=2)}{P(X=1)} = \dfrac{10p^2q^3}{5pq^4} = \dfrac{2p}{q}$.
   3. Numerically, $\dfrac{0.2048}{0.4096} = 0.5$, so $\dfrac{2p}{q}=0.5 \Rightarrow \dfrac{p}{q} = 0.25$.
   4. Since $p+q=1$, write $p = 0.25q$: $0.25q+q=1 \Rightarrow 1.25q=1 \Rightarrow q=0.8,\ p=0.2$. …

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10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.A dice is thrown twice. If getting a number greater than four is considered a success, the variance of the probability distribution of the number of successes is (A) $\dfrac{2}{9}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{4}{9}$

    ›Reveal solutionSolution

    The number of successes in two independent dice throws (success = number $>4$) follows Binomial$(2,\tfrac13)$, whose variance is $npq=\dfrac49$.

    Concept and Intuition

    Each throw is an independent Bernoulli trial with success probability $p=P(\text{die}>4)=P(5\text{ or }6)=\tfrac{2}{6}=\tfrac13$. The count of successes over $n=2$ independent trials is Binomial$(n,p)$, and the variance of a binomial variable has the closed form $npq$.

    Step-by-Step Solution

    1. $p=P(\text{success})=\dfrac{2}{6}=\dfrac13$ (numbers 5, 6 are $>4$), so $q=1-p=\dfrac23$.
    2. Number of successes $X\sim\text{Binomial}(n=2,\,p=\tfrac13)$. …

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11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.If the mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then $P(X=2) =$ (A) $\dfrac{7}{64}$ (B) $\dfrac{15}{64}$ (C) $\dfrac{21}{64}$ (D) $\dfrac{39}{64}$

    ›Reveal solutionSolution

    From mean $=np=4$ and variance $=npq=2$, we recover $p=\tfrac12$, $n=8$, and then compute $P(X=2)=\dfrac{7}{64}$ directly from the binomial pmf.

    Concept and Intuition

    A binomial distribution is fully determined by its parameters $n$ and $p$. Given the mean and variance, we can solve for $q=\mathrm{Var}/\mathrm{Mean}$ (since $\tfrac{npq}{np}=q$), then back out $p$ and $n$, after which any individual probability can be computed from the pmf formula.

    Step-by-Step Solution

    1. Mean: $np=4$. Variance: $npq=2$.
    2. Divide: $q=\dfrac{npq}{np}=\dfrac{2}{4}=\dfrac12$, so $p=1-q=\dfrac12$.
    3. From $np=4$: $n=\dfrac{4}{p}=\dfrac{4}{1/2}=8$. …

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12. CA Foundation 2026Set jan-20261 markMCQ
    Q.A quality control inspector finds that 20% of light bulbs are defective. If a batch of 5 light bulbs is tested, what is the probability that exactly 1 bulb is defective? (A) 0.4096 (B) 0.8026 (C) 0.2746 (D) 0.1296

    ›Reveal solutionSolution

    $P(X=1)=\binom{5}{1}(0.2)^1(0.8)^4=0.4096$.

    Step 1 — Set up the binomial model

    Each bulb is defective with probability $p=0.2$ independently; $n=5$ trials.

    Step 2 — Apply the binomial formula for $X=1$

    $P(X=1)=\binom{5}{1}p^{1}(1-p)^{4}=5\times0.2\times(0.8)^4.$

    Step 3 — Compute

    $(0.8)^4=0.4096,\qquad P=5\times0.2\times0.4096=1\times0.4096=0.4096.$ …

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