Probability (SAQs)

Maths-2A | 9. Probability – SAQs:
Welcome to SAQs in Chapter 9: Probability. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

Overview:

SAQ-1 : Suppose A and B are independent events with P(A)=0.6, P(B)=0.7 compute
i.P(A∩B)
ii.P(A∪B)
iii.P(B/A)
iv.P(A^c∩B^c)

Given that A,B are independent, hence

$$P(A \cap B) = P(A)P(B) = 0.6 \times 0.7 = 0.42$$

$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.6 + 0.7 – 0.42 = 1.3 – 0.42 = 0.88$$

$$P(B|A) = P(B) = 0.7$$

$$P(A^c \cap B^c) = P(A^c) \cdot P(B^c) = [1-P(A)][1-P(B)] = 0.4 \times 0.3 = 0.12$$

SAQ-2 : If A and B are independent events with P(A)=0.2, P(B)=0.5, find
i.P(A/B)
ii.P(B/A)
iii.P(A∩B)
iv.P(A∪B)

Given that A,B are independent hence

$$P(A|B) = P(A) = 0.2$$

$$P(B|A) = P(B) = 0.5$$

$$P(A \cap B) = P(A) \cdot P(B) = 0.2 \times 0.5 = 0.1$$

$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.2 + 0.5 – 0.1 = 0.6$$

SAQ-3 : If A,B are two events with P(A∪B)=0.65 and P(A∩B)=0.15, then find the value of P(A^c)+P(B^c)

We know $$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

$$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.65 + 0.15 = 0.8$$

$$P(A^c) + P(B^c) = [1-P(A)]+[1-P(B)]=2-[P(A)+P(B)]=2-0.8=1.2$$

SAQ-4 : If A,B are events with P(A)=0.5, P(B)=0.4 and P(A∩B)=0.3, find the probability that
i. A does not occur
ii. Neither A nor B occurs

Given $$P(A) = 0.5$$ $$P(B) = 0.4$$ and $$P(A \cap B) = 0.3$$

$$P(A^c) = 1-P(A) = 1-0.5 = 0.5$$

$$P(A^c \cap B^c) = P(A^c \cup B^c) = 1-P(A \cup B) = 1-[P(A)+P(B)-P(A \cap B)] = 1-(0.5+0.4-0.3) = 1-0.6 = 0.4$$

SAQ-5 : A problem in calculus is given two students A and B whose chances of solving it are 1/3,1/4 respectively. Find the probability of the problem being solved if both of them try independently

Let A,B denote the events of solving the problem by A,B respectively

$$P(A) = \frac{1}{3}$$ $$P(B) = \frac{1}{4}$$

$$P(A^c) = 1-P(A) = 1-\frac{1}{3}=\frac{2}{3}$$ $$P(B^c) = 1-P(B) = 1-\frac{1}{4}=\frac{3}{4}$$

$$P(A \cup B) = 1-P(A \cap B) = 1-P(A) \cdot P(B) = 1-(\frac{2}{3})(\frac{3}{4})=1-\frac{1}{2}=\frac{1}{2}$$

SAQ-6 : A speaks truth in 75% of the cases and B in 80% of the cases. What is the probability that their statements about an incident do not match

Let A,B denote the events of speaking truth by A,B respectively

$$P(A) = \frac{75}{100} = \frac{3}{4}$$

$$P(B) = \frac{80}{100} = \frac{4}{5}$$

$$P(A^c) = 1 – P(A) = 1 – \frac{3}{4} = \frac{1}{4}$$

$$P(B^c) = 1 – P(B) = 1 – \frac{4}{5} = \frac{1}{5}$$

Let E be the event that A and B contradict each other. Then,

$$P(E) = P\left((A \cap B^c) \cup (A^c \cap B)\right) = P(A \cap B^c) + P(A^c \cap B)$$

$$P(E) = P(A)P(B^c) + P(A^c)P(B) = \frac{3}{4} \cdot \frac{1}{5} + \frac{1}{4} \cdot \frac{4}{5} = \frac{7}{20}$$

SAQ-7 : The probability for a contractor to get a road contract is 2/3 and to get a building contract is 5/9. The probability to get atleast one contract is 4/5. Find the probability that he gets both the contracts

Let A be the event of getting road contract and B be the event of getting building contract

Given that $$P(A) = \frac{2}{3}$$ $$P(B) = \frac{5}{9}$$ $$P(A \cup B) = \frac{4}{5}$$

The probability that the contractor will get both contracts is

$$P(A \cap B) = P(A) + P(B) – P(A \cup B) = \frac{2}{3} + \frac{5}{9} – \frac{4}{5} = \frac{30}{45} + \frac{25}{45} – \frac{36}{45} = \frac{19}{45}$$

SAQ-8 : In a committee of 25 members, each member is proficient either in Mathematics or in Statistics or in both. If 19 of these are proficient in Mathematics, 16 in Statistics, Find the probability that a person selected from the committee is proficient in both

Let M, S denote the events of selecting a person who is proficient in Maths, Statistics respectively

Given that $$n(M \cup S) = 25$$ $$n(M) = 19$$ $$n(S) = 16$$ $$n(\mu) = 25$$

Now $$n(M \cap S) = n(M) + n(S) – n(M \cup S) = 19 + 16 – 25 = 10$$

$$P(M \cap S) = \frac{10}{25} = \frac{2}{5}$$

SAQ-9 : A,B,C are three horses in a race. The probability of A to win the race is twice that of B, and probability of B is twice that of C. What are probabilities of A,B,C to win the race?

Let A,B,C be the events of winning the race by the horses A,B,C respectively

Given that $$P(A) = 2P(B)$$ and $$P(B) = 2P(C)$$

Hence, $$P(A) = 2P(B) = 2[2P(C)] = 4P(C)$$

Now, $$P(A \cup B \cup C) = P(A) + P(B) + P(C)$$

$$P(S) = 4P(C) + 2P(C) + P(C)$$

$$1 = 7P(C)$$

$$P(C) = \frac{1}{7}$$

$$P(A) = 4P(C) = 4 \times \frac{1}{7} = \frac{4}{7}$$

$$P(B) = 2P(C) = 2 \times \frac{1}{7} = \frac{2}{7}$$

Therefore, $$P(A) = \frac{4}{7} P(B) = \frac{2}{7} P(C) = \frac{1}{7}$$

SAQ-10 : A,B,C are three news papers published from a city. 20% of the population read A, 16% read B, 14% read C, 8% read both A and B, 5% read both A and C, 4% read both B and C and 2% all the three. Find the percentage of the population who read atleast one news paper

Given that $$P(A) = \frac{20}{100} = 0.2$$ $$P(B) = \frac{16}{100} = 0.16$$ $$P(C) = \frac{14}{100} = 0.14$$

$$P(A \cap B) = \frac{8}{100} = 0.08$$

$$P(B \cap C) = \frac{4}{100} = 0.04$$

$$P(A \cap C) = \frac{5}{100} = 0.05$$

$$P(A \cap B \cap C) = \frac{2}{100} = 0.02$$

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C) – P(A \cap C) + P(A \cap B \cap C)$$

$$0.2 + 0.16 + 0.14 – 0.08 – 0.04 – 0.05 + 0.02 = 0.52 – 0.17 = 0.35$$

Percentage of population who read at least one newspaper = $$0.35 \times 100\% = 35\%$$

SAQ-11 : If A,B,C are three independent events such that P(A ∩B^c∩C^c)=1/4, P(A^c∩B∩C^c)=1/8,P(A^c∩B^c∩C^c)=1/4 then find P(A),P(B) and P(C)

Given that A,B,C are independent events

$$P(A \cap B^c \cap C^c) = \frac{1}{4} \Rightarrow P(A) \cdot P(B^c) \cdot P(C^c) = \frac{1}{4}…..(1)$$

$$P(A^c \cap B \cap C^c) = \frac{1}{8} \Rightarrow P(A^c) \cdot P(B) \cdot P(C^c) = \frac{1}{8}….(2)$$

$$P(A^c \cap B^c \cap C^c) = \frac{1}{4} \Rightarrow P(A^c) \cdot P(B^c) \cdot P(C^c) = \frac{1}{4}…..(3)$$

From (1)/(3) $$\Rightarrow \frac{P(A)}{P(A^c)} = \frac{1/4}{1/4} = 1$$

$$\Rightarrow \frac{P(A)}{1-P(A)} = 1$$

$$\Rightarrow P(A) = 1 – P(A)$$

$$\Rightarrow 2P(A) = 1$$

$$P(A) = \frac{1}{2}$$

From (2)/(3) $$\Rightarrow \frac{P(B)}{P(B^c)} = \frac{1/8}{1/4}$$

$$\Rightarrow \frac{P(B)}{1-P(B)} = \frac{1}{2}$$

$$\Rightarrow 2P(B) = 1 – P(B)$$

$$\Rightarrow 3P(B) = 1$$

$$P(B) = \frac{1}{3}$$

From (1), $$P(A) \cdot P(B^c) \cdot P(C^c) = \frac{1}{4}$$

$$\Rightarrow \left(\frac{1}{2}\right) \left(1-\frac{1}{3}\right) P(C^c) = \frac{1}{4}$$

$$\Rightarrow P(C^c) = \frac{1}{4} \times 2 \times \frac{3}{2} = \frac{3}{4}$$

$$P(C) = 1 – P(C^c) = 1 – \frac{3}{4} = \frac{1}{4}$$

Hence $$P(A) = \frac{1}{2} P(B) = \frac{1}{3} P(C) = \frac{1}{4}$$

SAQ-12 : Find the probability that a non-leap year contains
i.53 Sundays
ii.52 Sundays only

(i) Let E be the event of containing 53 Sundays in a non-leap year.

A non-leap year contains 365 days means 52 weeks and 1 day extra.

Here, $$S = {\text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}}$$

$$E = {\text{Sunday}}$$

Thus, $$n(S) = 7$$ and $$n(E) = 1$$ $$P(E) = \frac{n(E)}{n(S)} = \frac{1}{7}$$

(ii) If the remaining day is a non sunday then we get 52 sundays only

$$P(E’) = 1 – P(E) = 1 – \frac{1}{7} = \frac{6}{7}$$

SAQ-13 : State and prove multiplication theorem on probability

Statement: If A,B are 2 events of a random experiment such that $$P(A) \neq 0$$ $$P(B) \neq 0$$ then $$P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)$$

Proof: From the definition of conditional probability, $$P(B|A) = \frac{P(B \cap A)}{P(A)}$$

$$\Rightarrow P(B \cap A) = P(B|A) \cdot P(A)$$

$$\Rightarrow P(A \cap B) = P(A) \cdot P(B|A)$$

Again, from the definition of conditional probability, $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

$$\Rightarrow P(A \cap B) = P(A|B) \cdot P(B)$$

From (1) & (2) $$P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)$$

SAQ-14 : State conditional event and conditional probability

Conditional Event : Let A,B be two events of a sample space S. If B occurs only after the occurence of A, then the event of occurrence of B w.r.to A is called conditional event of B given A and it is denoted by B/A

Conditional Probability of B given A is $$P(B|A) = \frac{P(B \cap A)}{P(A)}$$

SAQ-15 : If A,B are two independent events in a sample space S, then prove that A ̅,B ̅ are independent

Given that A,B are independent events

$$P(A \cap B) = P(A)P(B)$$

$$P(A \cap B) = P(A \cup B) = 1 – P(A^c \cap B^c)$$

$$= 1 – [1 – P(A) – P(B) + P(A \cap B)]$$

$$= 1 – 1 + P(A) + P(B) – P(A)P(B)$$

$$= P(A) + P(B) – P(A)P(B)$$

For independent events,

$$P(A \cap B) = P(A)P(B)$$

$$P(A \cap B) = P(A) + P(B) – P(A)P(B)$$

This implies that $$P(A \cap B) = P(A)P(B)$$ confirms A, B are independent.