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 82 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.If $E$ and $F$ are two independent events such that $P(E) = \frac{3}{10}$, $P(E \cup F) = \frac{1}{2}$, then $P(E|F) - P(F|E)$ is equal to: (A) $\frac{2}{7}$ (B) $\frac{3}{35}$ (C) $\frac{1}{70}$ (D) $\frac{1}{7}$

   ›Reveal solutionSolution

   We use the property of independent events, $P(E \cap F) = P(E)P(F)$, along with the union formula to first find $P(F)$. Then, we use the fact that for independent events, $P(E|F) = P(E)$ and $P(F|E) = P(F)$, to calculate the required difference. The final result is $\boxed{\frac{1}{70}}$.

   The core of this problem lies in understanding how the concept of "independent events" simplifies probability calculations, especially when dealing with unions and conditional probabilities.

   When two events, $E$ and $F$, are independent, it means that the occurrence of one event does not affect the probability of the other event occurring. This has two crucial implications:

   1. Intersection Probability: The probability of both $E$ and $F$ happening, $P(E \cap F)$, is simply the product of their individual probabilities: $P(E \cap F) = P(E)P(F)$.
   2. Conditional Probability: The probability of $E$ happening given that $F$ has already happened, $P(E|F)$, is just the probability of $E$, because $F$'s occurrence doesn't change $E$'s likelihood. So, $P(E|F) = P(E)$. Similarly, $P(F|E) = P(F)$.

   We are given $P(E)$, $P(E \cup F)$, and that $E$ and $F$ are independent. Our strategy will be to first use the formula for the union of events, combined with the independence property, to find $P(F)$. Once we have $P(F)$, we can directly use the independence property to find $P(E|F)$ and $P(F|E)$, and then calculate their difference.

   1. Find $P(F)$ using the union formula and independence.

      The general formula for the probability of the union of two events is:

      $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

      Since $E$ and $F$ are independent, we can substitute $P(E \cap F)$ with $P(E)P(F)$:

      $P(E \cup F) = P(E) + P(F) - P(E)P(F)$

      Now, substitute the given values: $P(E) = \frac{3}{10}$ and $P(E \cup F) = \frac{1}{2}$.

      $\frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10} P(F)$

      To solve for $P(F)$, group the terms involving $P(F)$:

      $\frac{1}{2} = \frac{3}{10} + P(F) \left(1 - \frac{3}{10}\right)$

      $\frac{1}{2} = \frac{3}{10} + P(F) \left(\frac{7}{10}\right)$

      Subtract $\frac{3}{10}$ from both sides:

      $\frac{1}{2} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

      To subtract the fractions on the left, find a common denominator, which is $10$:

      $\frac{5}{10} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

      $\frac{2}{10} = P(F) \left(\frac{7}{10}\right)$

      Now, isolate $P(F)$ by multiplying both sides by $\frac{10}{7}$: …

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2. CBSE 2026Set V11 markMCQ
   Q.The probability of obtaining an even prime number on each die when a pair of dice is rolled

   (a) $\frac{1}{36}$

   (b) $\frac{1}{6}$

   (c) $\frac{1}{18}$

   (d) $\frac{1}{4}$

   ›Reveal solutionSolution

   The even prime is $2$; $P(2\text{ on each of two dice})=\tfrac{1}{6}\cdot\tfrac{1}{6}=\tfrac{1}{36}$; answer (a).

   The only even prime number is $2$. For one die, $P(\text{show }2)=\dfrac{1}{6}$. The two dice are independent, so …

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3. CBSE 2026Set V11 markMCQ
   Q.If $A$ and $B$ are independent events with $P(A)=0.3$ and $P(B)=0.4$ then $P(A\cap B)$

   (a) $1.2$

   (b) $0.12$

   (c) $0.7$

   (d) $\frac{3}{4}$

   ›Reveal solutionSolution

   Independence gives $P(A\cap B)=P(A)P(B)=0.12$; answer (b).

   For independent events $A$ and $B$,

   $P(A\cap B)=P(A)\cdot P(B)=0.3\times0.4=0.12.$ …

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4. CBSE 2026Set A1 markMCQ
   Q.If $A$, $B$ and $C$ are three independent events then $P(ABC) =$

   (a) $P(A) + P(B) + P(C)$

   (b) $P(A) - P(B) - P(C)$

   (c) $P(A) \cdot P(B) \cdot P(C)$

   (d) None of these

   ›Reveal solutionSolution

   For independent events, $P(A\cap B\cap C) = P(A)P(B)P(C)$.

   By definition, events $A$, $B$, $C$ are (mutually) independent when the probability of their joint occurrence equals the product of their individual probabilities:

   $P(ABC) = P(A)\cdot P(B)\cdot P(C).$

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.In a box containing 100 bulbs, 10 are defective. Write the probability that out of a sample of 5 bulbs, none is defective.

   (a) $10^{-1}$

   (b) $\left(\frac{1}{2}\right)^5$

   (c) $\left(\frac{9}{10}\right)^5$

   (d) $\frac{9}{10}$

   ›Reveal solutionSolution

   Treating each draw as an independent trial with probability $\frac{9}{10}$ of being non-defective, the probability that all $5$ are non-defective is $\left(\frac{9}{10}\right)^5$.

   Out of $100$ bulbs, $10$ are defective, so the probability a randomly chosen bulb is not defective is

   $P(\text{good})=\frac{100-10}{100}=\frac{90}{100}=\frac{9}{10}$

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If A and B are independent events and $P(A) = 0.3$ and $P(B) = 0.4$, then the value of $P(A \cup B)$ will be

   (a) $0.58$

   (b) $0.70$

   (c) $0.12$

   (d) $0.10$

   ›Reveal solutionSolution

   For independent events, $P(A\cap B)=P(A)P(B)$, and the addition rule gives $P(A\cup B)$.

   $P(A\cap B) = 0.3\times0.4=0.12$ (independence).

   …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.The probability of obtaining an even prime number on each dice, when a pair of dice is rolled, is:

   (a) 0

   (b) $\dfrac{1}{3}$

   (c) $\dfrac{1}{12}$

   (d) $\dfrac{1}{36}$

   ›Reveal solutionSolution

   The only even prime number is 2, so we need a 2 on each die.

   Among $\{1,2,3,4,5,6\}$, the only even prime is $2$.

   $P(\text{2 on one die})=\dfrac{1}{6}$

   …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.Ajay and Meera are contesting for two vacancies in a company. Probability of selection of Ajay is 7/9 and that of Meera is 4/7. What is the probability that both will be rejected?

   (a) 61/63

   (b) 6/63

   (c) 41/63

   (d) 28/63

   ›Reveal solutionSolution

   The rejection probabilities of each candidate are complements of their selection probabilities; since the two events are independent, multiply them.

   $P(\text{Ajay selected}) = \dfrac79 \implies P(\text{Ajay rejected}) = 1-\dfrac79 = \dfrac29$

   $P(\text{Meera selected}) = \dfrac47 \implies P(\text{Meera rejected}) = 1-\dfrac47 = \dfrac37$

   …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): Two events A and B are such that P(A) = 1/4, P(B) = 1/2 and P(A∩B) = 1/8 then two events A and B are independent. Reason (R): Two events are independent if the probability of occurrence of one does not affect the probability of occurrence of other and P(A∩B) = P(A) + P(B) − P(A∪B)

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A)

   (c) Assertion (A) is true but Reason (R) is false.

   (d) Both Assertion (A) and Reason (R) are false.

   ›Reveal solutionSolution

   The independence check in A is numerically correct, but R states the wrong criterion — it gives the addition-rule identity (true for ANY two events), not the actual independence condition $P(A\cap B)=P(A)\cdot P(B)$.

   Checking Assertion (A): Independence requires $P(A\cap B) = P(A)\cdot P(B)$.

   $P(A)\cdot P(B) = \frac14 \times \frac12 = \frac18,$

   which equals the given $P(A\cap B)=1/8$. So A and B are independent — A is true.

   Checking Reason (R): R correctly describes independence in words ("occurrence of one does not affect the other"), but then states the test as

   $P(A\cap B) = P(A)+P(B)-P(A\cup B).$ …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.Let $E$ and $F$ be events with $P(E) = \dfrac{1}{3}$, $P(F) = \dfrac{1}{2}$ and $P(E \cap F) = \dfrac{1}{6}$. Then

    (a) $E$ and $F$ are independent events

    (b) $E$ and $F$ are mutually exclusive events

    (c) $E$ and $F$ are disjoint events

    (d) None of the above

    ›Reveal solutionSolution

    Two events are independent exactly when $P(E\cap F)=P(E)\cdot P(F)$; check whether the given numbers satisfy this.

    Given $P(E)=\dfrac13$, $P(F)=\dfrac12$, $P(E\cap F)=\dfrac16$.

    Test for independence:

    $P(E)\cdot P(F) = \frac13\times\frac12 = \frac16$

    This equals the given $P(E\cap F)=\dfrac16$. Since $P(E\cap F)=P(E)P(F)$, $E$ and $F$ are independent.

    …

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11. CBSE 2026Set ANNUAL1 mark
    Q.If A and B are two independent events with P(A) = 1/2 and P(B) = 1/3, then find P(A ∪ B).

    ›Reveal solutionSolution

    For independent events, $P(A\cap B)=P(A)P(B)$; then apply the addition rule.

    …

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12. CBSE 2026Set ANNUAL1 mark
    Q.Write the answer in one word/sentence: If $E$ and $F$ are independent events then write the value of $P(E \cap F)$.

    ›Reveal solutionSolution

    For independent events, the probability of the intersection is the product.

    …

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