Class 12 Probability Mock Test (CBSE)

Topics in Probability

• Conditional probability
• Properties of conditional probability
• Multiplication theorem on probability
• Independent events
• Theorem of total probability
• Bayes' theorem

Probability practice questions with answers

1. Q1Multiple choice·1 mark
   Bag I contains $3$ red and $2$ black balls, while Bag II contains $1$ red and $4$ black balls. One bag is chosen at random and a ball is drawn from it, which is found to be red. The probability that it was drawn from Bag I is

   1. (a)$1/4$
   2. (b)$2/5$
   3. (c)$3/5$
   4. (d)$3/4$
   Show answer and step marks

   1. Correct option: (d) $3/4$. Let $E_1$ and $E_2$ be events of choosing Bag I and Bag II respectively, and $R$ be the event of drawing a red ball. $P(E_1) = 1/2$, $P(E_2) = 1/2$, $P(R|E_1) = 3/5$, $P(R|E_2) = 1/5$. By Bayes' theorem, $P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)} = \frac{(1/2)(3/5)}{(1/2)(3/5) + (1/2)(1/5)} = \frac{3/5}{4/5} = \frac{3}{4}$.+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   For two events $A$ and $B$ with $P(B) > 0$, it is given that $P(A \mid B) = 0.35$. The value of $P(A' \mid B)$ is

   1. (a)$0.35$
   2. (b)$0.65$
   3. (c)$0.1225$
   4. (d)$1.35$
   Show answer and step marks

   1. Correct option: (b) $0.65$. Since $P(A' \mid B) = 1 - P(A \mid B) = 1 - 0.35 = 0.65$.+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   Let $E_1$ and $E_2$ be two mutually exclusive and exhaustive events associated with a random experiment such that $P(E_1) = 0.4$ and $P(E_2) = 0.6$. If $A$ is any event such that $P(A|E_1) = 0.2$ and $P(A|E_2) = 0.3$, then $P(A)$ is equal to

   1. (a)$0.26$
   2. (b)$0.18$
   3. (c)$0.50$
   4. (d)$0.12$
   Show answer and step marks

   1. Correct option: (a) $0.26$. Using the theorem of total probability, $P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) = (0.4)(0.2) + (0.6)(0.3) = 0.08 + 0.18 = 0.26$.+1

   Answer: (a)
4. Q4Multiple choice·1 mark
   Events $A$ and $B$ are such that $P(A) = 0.8$ and $P(B \mid A) = 0.25$. Using the multiplication theorem on probability, the value of $P(A \cap B)$ is

   1. (a)$0.25$
   2. (b)$0.8$
   3. (c)$1.05$
   4. (d)$0.2$
   Show answer and step marks

   1. Correct option: (d) $0.2$. By the multiplication theorem, $P(A \cap B) = P(A) \cdot P(B \mid A) = 0.8 \times 0.25 = 0.2$.+1

   Answer: (d)
5. Q5Short answer·3 marks
   A fair die is thrown once. Let $A$ be the event that the number obtained is even and $B$ be the event that the number obtained is a multiple of $3$. (a) Write the condition that must hold for the events $A$ and $B$ to be independent. [1] (b) Find $P(A)$, $P(B)$ and $P(A \cap B)$. [1] (c) Are the events $A$ and $B$ independent? Justify your answer with a calculation. [1]
   Show answer and step marks

   1. Condition for independence: $P(A \cap B)=P(A)\cdot P(B)$+1
   2. Sample space $\{1,2,3,4,5,6\}$: $P(A)=\frac{3}{6}=\frac{1}{2}$, $P(B)=\frac{2}{6}=\frac{1}{3}$, $P(A\cap B)=P(\{6\})=\frac{1}{6}$+1
   3. $P(A)\cdot P(B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}=P(A\cap B)$, so $A$ and $B$ are independent+1

   Answer: (a) $P(A\cap B)=P(A)\cdot P(B)$; (b) $P(A)=\frac{1}{2}$, $P(B)=\frac{1}{3}$, $P(A\cap B)=\frac{1}{6}$; (c) Yes, independent, since $\frac{1}{6}=\frac{1}{2}\times\frac{1}{3}$
6. Q6Short answer·3 marks
   A courier company sends parcels through three routes $R_1$, $R_2$ and $R_3$. It sends $50\%$ of its parcels by route $R_1$, $30\%$ by route $R_2$ and $20\%$ by route $R_3$. Of the parcels sent by $R_1$, $4\%$ arrive late; of those sent by $R_2$, $6\%$ arrive late; and of those sent by $R_3$, $10\%$ arrive late. (a) Explain why the events $R_1$, $R_2$ and $R_3$ form a partition of the sample space of all parcels. [1] (b) Let $L$ be the event that a parcel arrives late. Using the theorem of total probability, find $P(L)$. [2]
   Show answer and step marks

   1. Each parcel is sent by exactly one route, so the routes are mutually exclusive; the percentages total $50+30+20=100\%$, so the routes cover all parcels and are exhaustive+1
   2. $P(L)=P(R_1)P(L\mid R_1)+P(R_2)P(L\mid R_2)+P(R_3)P(L\mid R_3)$+1
   3. $=0.5(0.04)+0.3(0.06)+0.2(0.10)=0.02+0.018+0.02=0.058$+1

   Answer: (a) The routes are mutually exclusive and exhaustive; (b) $P(L)=0.058$

More Class 12 Maths mock tests