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.


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

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


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

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

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


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


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

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

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

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

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

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


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

White BlackRed
I123
II211
III453

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

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

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

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

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


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

WhiteBlackRed
B1212
B2324
B3432

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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


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 $$\frac{1}{3} \leq p \leq \frac{1}{2}$$