Probability (LAQs)

Maths-2A | 9. Probability – LAQs:
Welcome to LAQs in Chapter 9: Probability. This page contains the most Important FAQs for Long 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.

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LAQ-1 : State and prove addition theorem on probability. (or) If E₁,E₂ are any two events of a random experiment and P is probability function then prove that P(E₁∪E₂)=P(E₁)+P(E₂)-P(E₁∩E₂)

Statement : If E₁,E₂ are the 2 events of a sample space S then $$P(E_1 \cup E_2) = P(E_1) + P(E_2) – P(E_1 \cap E_2)$$

Proof:

Case (i) : When $$E_1 \cap E_2 = \emptyset$$

$$E_1 \cap E_2 = \emptyset \Rightarrow P(E_1 \cap E_2) = 0$$

$$P(E_1 \cup E_2) = P(E_1) + P(E_2)$$

$$= P(E_1) + P(E_2) – 0 = P(E_1) + P(E_2) – P(E_1 \cap E_2)$$

Case (ii) : When $$E_1 \cap E_2 \neq \emptyset$$

$$E_1 \cup E_2$$ is the union of disjoint $$(E_1 – E_2)$$ and $$E_2$$

$$P(E_1 \cup E_2) = P[(E_1 – E_2) \cup E_2] = P(E_1 – E_2) + P(E_2)….(1)$$

E₁ is the union of disjoint sets $$(E_1 – E_2)$$ $$E_1 \cap E_2$$

$$P(E_1) = P[(E_1 – E_2) \cup (E_1 \cap E_2)] = P(E_1 – E_2) + P(E_1 \cap E_2)$$

$$\Rightarrow P(E_1 – E_2) = P(E_1) – P(E_1 \cap E_2)$$

From (1), $$P(E_1 \cup E_2) = [P(E_1) – P(E_1 \cap E_2)] + P(E_2)$$

$$= P(E_1) + P(E_2) – P(E_1 \cap E_2)$$

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LAQ-2 : Find the probability of drawing an Ace or a spade from a well shuffled pack of 52 playing cards

Number of ways of selecting a card from 52 cards is $$n(S) = {52 \choose 1} = 52$$

Let E₁ be the event of drawing a spade card ⇒ $$n(E_1) = {13 \choose 1} = 13$$

Let E₂ be the event of drawing Ace card ⇒ $$n(E_2) = {4 \choose 1} = 4$$

Also, $$n(E_1 \cap E_2) = 1$$

$$P(E_1 \cup E_2) = P(E_1) + P(E_2) – P(E_1 \cap E_2)$$

$$= \frac{13}{52} + \frac{4}{52} – \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$

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LAQ-3 : State and Prove Baye’s theorem on probability

Statement : If $$E_1, E_2, \ldots, E_n$$ are mutually exclusive and exhaustive events in a sample space S and A is any event intersecting with any E_i such that $$P(A) \neq 0$$ then $$P(E_k | A) = \frac{P(E_k)P(A | E_k)}{\sum_{i=1}^{n} P(E_i)P(A | E_i)}$$

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

Given that $$E_1, E_2, \ldots, E_n$$ are mutually exclusive and exhaustive events in a sample space S

⇒ $$\bigcup_{i=1}^{n} E_i = S$$ and $$A \cap E_1, A \cap E_2, \ldots, A \cap E_n$$ are mutually disjoint ⇒ $$A \cap E_i = \emptyset$$

Now $$P(A) = P(S \cap A) = P\left(\left(\bigcup_{i=1}^{n} E_i\right) \cap A\right) = P\left(\bigcup_{i=1}^{n}(E_i \cap A)\right) = \sum_{i=1}^{n} P(E_i \cap A)$$

From (1), $$P(E_k | A) = \frac{P(E_k)P(A | E_k)}{\sum_{i=1}^{n} P(E_i)P(A | E_i)}$$

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LAQ-4 : Suppose that an urn B₁ contains 2 white and 3 black balls and another urn B₂ contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was B₁

Let E₁,E₂ denote the events of selecting urn B₁ and urn B₂ respectively and B be the event of drawing a black ball

Then $$P(E_1) = P(E_2) = \frac{1}{2}$$ and $$P(B | E_1) = \frac{3}{5}$$ $$P(B | E_2) = \frac{4}{7}$$

By Baye’s theorem the required probability is

$$P(E_1 | B) = \frac{P(E_1)P(B | E_1)}{P(E_1)P(B | E_1) + P(E_2)P(B | E_2)}$$

$$= \frac{\frac{1}{2} \times \frac{3}{5}}{\left(\frac{1}{2} \times \frac{3}{5}\right) + \left(\frac{1}{2} \times \frac{4}{7}\right)}$$

$$= \frac{3/10}{3/10 + 2/7}$$

$$= \frac{3/10 \times 70}{41}$$

$$= \frac{21}{41}$$

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LAQ-5 : Three boxes numbered I, II, III contains the balls as follows:

	White	Black	Red
I	1	2	3
II	2	1	1
III	4	5	3

Let B₁,B₂,B₃ be the events of selecting boxes B₁,B₂,B₃ and R be the event of getting drawing a red ball

$$P(B_1) = \frac{1}{3}, P(B_2) = \frac{1}{3}, P(B_3) = \frac{1}{3}$$

$$P(R | B_1) = \frac{3}{6} = \frac{1}{2}, P(R | B_2) = \frac{1}{4}, P(R | B_3) = \frac{3}{12} = \frac{1}{4}$$

By Baye’s theorem the required probability os

$$P(B_2 | R) = \frac{P(B_2)P(R | B_2)}{P(B_1)P(R | B_1) + P(B_2)P(R | B_2) + P(B_3)P(R | B_3)}$$

$$= \frac{\frac{1}{3} \times \frac{1}{4}}{\left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{1}{4}\right)}$$

$$= \frac{\frac{1}{4}}{\frac{1}{2} + \frac{1}{4} + \frac{1}{4}}$$

$$P(B_2 | R) = \frac{1}{4}$$

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LAQ-6 : Three boxes B₁, B₂ and B₃ contain balls with different colours are shown here
A die is thrown and B₁ is chosen if either 1 or 2 turns up; B₂ is chosen if 3 or 4 turns up and B₃ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball drawn is of red colour, what is the probability that it comes from box B₂

	White	Black	Red
B1	2	1	2
B2	3	2	4
B3	4	3	2

Let B₁, B₂, B₃ be the events of selecting boxes B₁, B₂, B₃ and R be the event of getting drawing a red ball

$$P(B_1) = \frac{1}{3}$$ $$P(B_2) = \frac{1}{3}$$ $$P(B_3) = \frac{1}{3}$$

$$P(R | B_1) = \frac{2}{5}$$ $$P(R | B_2) = \frac{4}{9}$$ $$P(R | B_3) = \frac{2}{9}$$

By Baye’s theorem the required probability is

$$P(B_2 | R) = \frac{P(B_2)P(R | B_2)}{P(B_1)P(R | B_1) + P(B_2)P(R | B_2) + P(B_3)P(R | B_3)}$$

$$= \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{1}{3} \times \frac{2}{5} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{2}{9}}$$

$$= \frac{4/9}{(2/5 + 4/9 + 2/9) \times 1/3}$$

$$= \frac{4/9}{\left(\frac{18 + 20 + 10}{45}\right) \times \frac{1}{3}}$$

$$= \frac{20}{48} = \frac{5}{12}$$

$$P(B_2 | R) = \frac{5}{12}$$

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LAQ-7 : Three urns have the following composition of balls
Urn I: 1 white, 2 black Urn
Urn II: 2 white, 1 black
Urn III: 2 white, 2 black
One of the urns is selected at random and a ball is drawn. It turns out to be white. Find the probability that it came from urn III

Let U₁, U₂, U₃ be the events of selecting urn I, urn II, urn III and W be the event of getting drawing a white ball

$$P(U_1) = \frac{1}{3}$$ $$P(U_2) = \frac{1}{3}$$ $$P(U_3) = \frac{1}{3}$$

By Baye’s theorem the required probability is

$$P(U_3 | W) = \frac{P(U_3)P(W | U_3)}{P(U_1)P(W | U_1) + P(U_2)P(W | U_2) + P(U_3)P(W | U_3)}$$

$$= \frac{\frac{1}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{2}}$$

$$= \frac{\frac{1}{6}}{\frac{1}{9} + \frac{2}{9} + \frac{1}{6}}$$

$$= \frac{1/6}{\frac{3}{9} + \frac{1}{2}}$$

$$= \frac{1/6}{\frac{1}{3} + \frac{1}{2}}$$

$$= \frac{1/6}{\frac{2}{6} + \frac{3}{6}}$$

$$= \frac{1/6}{\frac{5}{6}}$$

$$P(U_3 | W) = \frac{1}{5}$$

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LAQ-8 : The probabilities of three mutually exclusive events are respectively given as 1+3p)/3, 1-p/4, 1-2p/2. Prove that 1/3≤p≤1/2

The probability of any event lies between 0 and 1.

Hence we have $$0 \leq \frac{1+3p}{3} \leq 1

Hence we have $$0 \leq \frac{1+3p}{3} \leq 1$$ $$0 \leq \frac{1-p}{4} \leq 1$$ $$0 \leq \frac{1-2p}{2} \leq 1$$

Now, $$0 \leq \frac{1+3p}{3} \leq 1 \Rightarrow 0 \leq 1+3p \leq 3 \Rightarrow -1 \leq 3p \leq 2 \Rightarrow -\frac{1}{3} \leq p \leq \frac{2}{3}$$

$$0 \leq \frac{1-p}{4} \leq 1 \Rightarrow 0 \leq 1-p \leq 4 \Rightarrow 1 \leq -p \leq 3 \Rightarrow 1 \geq p \geq -3 \Rightarrow -3 \leq p \leq 1$$

$$0 \leq \frac{1-2p}{2} \leq 1 \Rightarrow 0 \leq 1-2p \leq 2 \Rightarrow -1 \leq -2p \leq 1 \Rightarrow 1 \geq 2p \geq -1 \Rightarrow -\frac{1}{2} \leq p \leq \frac{1}{2}$$

The given 3 events are mutually exclusive of the same random experiment

$$\Rightarrow 0 \leq \frac{1+3p}{3} + \frac{1-p}{4} + \frac{1-2p}{2} \leq 1 \Rightarrow 0 \leq \frac{4(1+3p)+3(1-p)+6(1-2p)}{12} \leq 1 \Rightarrow 0 \leq 4 + 12p + 3 – 3p + 6 – 12p \leq 12$$

$$\Rightarrow 0 \leq 13 – 3p \leq 12 \Rightarrow -13 \leq -3p \leq -1 \Rightarrow 13 \geq 3p \geq 1 \Rightarrow \frac{1}{3} \leq p \leq \frac{13}{3}$$

From (1), (2), (3), (4) the common intersection region is $$\frac{1}{3} \leq p \leq \frac{1}{2}$$