Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type

Question 1.

If one ticket is randomly selected from ticket numbers 1 to 30, then find the probability that the number on the ticket

is

I. a multiple of 5 or 7

II. a multiple of 3 or 5 [Mar. ’08]

Solution:

Let, S be the sample space.

No. of ways of drawing one ticket from 30 ticket

∴ n(S) = \({ }^{30} \mathrm{C}_1\) = 30

I) Let K be the event of getting a multiple of 5 then A = {5, 10, 15, 20, 25, 30}

∴ n(A) = 6

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{30}=\frac{1}{5}\)

Let B be the event of getting a multiple of 7 then

B = {7, 14, 21, 28}

∴ n(B) = 4

∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{30}=\frac{2}{15}\)

A ∩ R = getting a multiple of 5 and 9 = Φ

∴ n(A ∩ B) = 0

P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{0}{30}\) = 0

The probability that the number on the ticket is a multiple of 5 or 7.

According to addition theorem on probability,

P(A ∪ B) = P(A) + P(ß) – P(A ∩ B)

= \(\frac{1}{5}+\frac{2}{15}\) – 0

= \(\frac{3+2}{15}=\frac{5}{15}=\frac{1}{3}\).

II) Let A be the event of getting a multiple of 3 then

A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

∴ n(A) = 10

∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{10}{30}=\frac{1}{3}\)

Let B be the event of getting a multiple of 5 then

B = {5, 10, 15, 20, 25, 30}

∴ n(B) = 6

∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{6}{30}=\frac{1}{5}\)

A ∩ B is the event getting a multiple of 5 and 3 then A ∩ B = {15, 30}

∴ n(A ∩ B) = 2

∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{30}=\frac{1}{15}\)

The probability that the number on the ticket is a multiple of 3 or 5. According to addition theorem on probability.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\)

= \(\frac{5+3-1}{15}=\frac{7}{15}\).

Question 2.

The probabilities of three events A, B, C are such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09 and P(A ∪ B ∪ C) = 0.75. Show that P(B ∩ C) lies in the interval [0.23, 0.48]. [Board Paper]

Solution:

Given that,

P(A) = 0.3, P(B) = 0.4, P(C) = 0.8

P(A ∩ B) = 0.08, P(A ∩ C) = 0.28

P(A ∩ B ∩ C) = 0.09, and P(A ∪ B ∪ C) ≥ 0.75

Now,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

= 0.3 + 0.4 + 0.8 – 0.08 – P(B ∩ C) – 0.28 + 0.09

= – 0.27 + 1.5 – P(B ∩ C)

= 1.23 – P(B ∩ C)

Since, P(A ∪ B ∪ C) ≥ 0.75

⇒ 0.75 ≤ P(A ∪ B ∪ C) ≤ 1

⇒ 0.75 ≤ 1.23 – P(B ∩ C) ≤ 1

⇒ 0.75 – 1.23 ≤ P(B ∩ C) ≤ 1 – 1.23

⇒ -0.48 ≤ – P(B ∩ C) ≤ – 0.23

⇒ 0.48 ≥ P(B ∩ C) ≥ 0.23

⇒ 0.23 ≤ P(B ∩ C) ≤ 0.48

∴ P(B ∩ C) c [0.23, 0.48]

∴ P(B ∩ C) lies in the interval [0.23, 0.48].

Question 3.

The probabilities of 3 mutually exclisive events are respectively given as \(\frac{1+3 \mathbf{P}}{3}, \frac{1-\mathbf{P}}{\mathbf{4}}, \frac{\mathbf{1 – 2 P}}{\mathbf{2}}\). Prove that \(\frac{1}{3}\) ≤ p ≤ \(\frac{1}{2}\)

Solution:

Suppose, A, B, C are exclusive event such that

P(A) = \(\frac{1+3 p}{3}\)

P(B) = \(\frac{1-p}{3}\)

P(C) = \(\frac{1-2 p}{2}\)

We know that

0 ≤ P(A) ≤ 1

0 ≤ \(\frac{1+3 \mathrm{p}}{3}\) ≤ 1

0 ≤ 1 + 3p ≤ 3

0 ≤ 1 ≤ 3p ≤ 3 – 1

– 1 ≤ 3p ≤ 2

\(\frac{-1}{3} \leq p \leq \frac{2}{3}\) …………….(1)

0 ≤ P(B) ≤ 1

0 ≤ \(\frac{1-p}{4}\) ≤ 1

0 ≤ 1 – p ≤ 4

0 – 1 ≤ 1 – p – 1 ≤ 4 – 1

– 1 ≤ – p ≤ 3

1 ≥ p ≥ – 3

– 3 ≤ p ≤ 1 ……………(2)

0 ≤ \(\frac{1-2 p}{2}\) ≤ 1

0 ≤ 1 – 2p ≤ 2

o – 1 ≤ 1 – 2p – 1 ≤ 2 – 1

– 1 ≤ – 2p ≤ 1

\(+\frac{1}{2}\) p ≥ \(\frac{1}{2}\)

\(\frac{1}{2}\) ≤ p ≤ \(\frac{1}{2}\)

Since, A, B, C are mutually exclusive events then 0 ≤ P(A ∪ B ∪ C) ≤ 1

0 ≤ P(A) + P(B) + P(C) ≤ 1

0 ≤ \(\frac{1+3 p}{3}+\frac{1-p}{4}+\frac{1-2 p}{2}\) ≤ 1

0 ≤ \(\frac{4+12 p+3-3 p+6-12 p}{12}\) ≤ 1

0 ≤ \(\frac{13-3 p}{12}\) ≤ 1

0 ≤ 13 – 3p ≤ 12

0 – 13 ≤ – 3p ≤ 12 – 13

– 13 ≤ – 3p ≤ – 1

Question 4.

State and prove addition theorem on probability [May ‘14, ‘12, ‘09, ‘08, ’07, ‘06, ‘05, Mar. ’14. ‘11. ‘07; AP – Mar. ‘18. ‘17; TS – Mar.’18. May ‘16]

Solution:

Addition theorem on probability:

If A, B are two events in a sample space, S then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Proof:

In a sample space

A = {a_{1}, a_{2}, …………………. a_{1}, a_{1+1}, ……………. a_{m})

B = {a_{1+1} ……… am, a_{m+1} …………………… a_{n}}

A ∩ B = {a_{1+1}, ………………., a_{m}}

A ∪ B = {a_{1}, a_{2}, ……….. a_{1}, a_{1+1}, a_{m}, a_{m+1}, …………..a_{m}}

According to definition of probability

P(A) = \(\sum_{i=1}^m\) P(a_{i})

P(B) = \(\sum_{\mathbf{i}=l+1}^{\mathrm{n}}\) P(a_{i})

P(C) = \(\sum_{\mathbf{i}=l+1}^{\mathrm{n}}\) P(a_{i})

Question 5.

If A, B, C are 3 independent events of an experiment such that \(P\left(A \cap B^C \cap C^C\right)=\frac{1}{4}\), \(P\left(A^C \cap B \cap C^C\right)=\frac{1}{8}\), \(P\left(A^C \cap B \cap C^C\right)=\frac{1}{4}\) then P(A), P(B) and P(C). [Mar. ’10, AP – May 2016; TS – Mar. 2015]

Solution:

Given that,

A, B, C are three independent events then \(A^{\mathrm{C}}, \mathrm{B}^{\mathrm{C}}, \mathrm{C}^{\mathrm{C}}\) are also independent events.

Question 6.

State and prove multiplication theorem on probability. [TS – May ’15, May ’10, Mar. ’04]

Solution:

Multiplication theorem of probability: (or) Theorem of compound probability:

Let A, B be two events in a sample s pace, S such that P(A) ≠ 0, P(B) ≠ 0 then

I) P(A ∩ B) = P(A). \(\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\)

II) P(A ∩ B) = P(B). \(\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)\)

Proof:

Let, n(A), n(B), n(A ∩ B), n(S) be the number of sample points in A, B, A ∩ B, S respectively.

Then,

Question 7.

State and prove Baye’s theorem on probability,

Statement: If A_{1}, A_{2}, ………….., A_{n}, are mutually exclusive and exhaustive events in a sample space, S such that P(A_{1}) > 0 for i = 1, 2, …, n and E is any event with P(E) > 0 then \(P\left(\frac{A_K}{E}\right)=\frac{P\left(A_K\right) \cdot P\left(\frac{E}{A_K}\right)}{\sum_{i=1}^n P\left(A_i\right) P\left(\frac{E}{A_i}\right)}\) for K = 1, 2, ………….., n. [Mar. 12, ‘09, May ‘05 AP – Mar. May, ’15, ’16, TS – Mar. ‘17, ‘16, AP – Mar. 2019]

Solution:

Proof :

Since, A_{1}, A_{2}, …………….. A_{n} are mutually exclusive and exhaustive events in a sample space, S it follows that \(\bigcup_{i=1}^n\) A_{i} = S and

A_{1}, A_{2}, ………….. A_{n} are mutually disjoint.

Now,

E ∩ A_{1}, E ∩ A_{2}, ……………….. E ∩ A_{n} are mutually disjoint.

∴ P(E) = P(E ∩ S)

Question 8.

Suppose that an urn B_{1} contains two white and 3 black balls and another urn B_{2} 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 pobability that the urn choosen was B_{1}.

Solution:

Let E_{1}, E_{2} denote the events of selecting mens B_{1} and B_{2} respectively then

P(E_{1}) = \(\frac{1}{2}\),

P(E_{2}) = \(\frac{1}{2}\)

Let B denote the event that the ball choosen from the selected men is black then

Question 9.

Three boxes B_{1}, B_{2} and B_{3} contain balls with different colours as shown below.

A die is thrown. B_{1} is chosen if either 1 or 2 turns up. B_{2} is chosen If 3 or 4 turns up and B_{3}, is chosen if 5 or 6 turns up. Having chosen a box In this way, a ball is choosen at random from this box. If the ball drawn Is found to be red, find the probability that it is drawn fron box B_{2}.

Solution:

Let E_{1}, E_{2}, E_{3} denote the events of selecting boxes B_{1}, B_{2}, B_{3} respectively.

∴ P(E_{1}) = \(\frac{2}{6}=\frac{1}{3}\)

P(E_{2}) = \(\frac{2}{6}=\frac{1}{3}\)

P(E_{3}) = \(\frac{2}{6}=\frac{1}{3}\)

Let f denote the event that the ball choosen from the selected box is red.

Question 10.

An urn contains ‘w’ white halls and ‘b’ black balls. Two players Q and R alternatively draw a ball with replacement from the urn. The player that draws a white ball first wins the game. If Q begins the game, find the probability of his winning the game.

Solution:

Let w denote the event of drawing a white ball in any draw then

P(w) = \(\frac{w_{C_1}}{(w+b)_{C_1}}=\frac{w}{w+b}\)

Let B denote the event of drawing a black ball in any draw then

P(B) = \(\frac{\mathrm{b}_{\mathrm{C}_1}}{(\mathrm{w}+\mathrm{b})_{\mathrm{C}_1}}=\frac{\mathrm{b}}{\mathrm{w}+\mathrm{b}}\)

The probability of Q wins the game.

= P(w ∪ BBw ∪ BBBBw ∪ ……………. )

= P(w) + P(BBw) + P(BBBB)w) + ……………..

= P(w) + P(B)P(B)P(w) + P(B) P(B) P(B) P(B) P(w) + …………….

= P(w) [1 [P(B)^{2}+ [P(B)]^{4} + ……………….]

Question 11.

Three urns have the following composition of balls:

urn I : 1 white, 2 black

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 of two be white. Find the probability that It came

from urn III. [AP – Mar. 2017] [May ’13]

Solution:

Let, A_{1}, A_{2}, A_{3} be the events of selecting urn – I, urn – II, urn – III respectively then

P(A_{1}) = \(\frac{1}{3}\),

P(A_{2}) = \(\frac{1}{3}\),

P(A_{3}) = \(\frac{1}{3}\)

Now A_{1}, A_{2}, A_{3} are mutually exclusive and exhaustive events.

Let E be the event of drawing a white ball from the selected urn.

Question 12.

A person is known to speak truth 2 out of 3 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.

Solution:

Let, A be the event that 1 occurs when a die is thrown.

∴ P(A) = \(\frac{1}{6}\)

Let, E be the event that the man reports that it is 1. Since, the man speaks the truth 2 out of 3 times.

Question 13.

Three boxes numbered I, II, III contain the balls as follows:

One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that It Is from Box – II. [TS. Mar. 2019]

Solution:

Let A_{1}, A_{2}, A_{3} be the events of drawing a ball from the box numbered I, II, III respectively and E be the event of drawing a red ball from the selected box.

P(A_{1}) = \(\frac{1}{3}\),

P(A_{2}) = \(\frac{1}{3}\),

P(A_{3}) = \(\frac{1}{3}\) and

A_{1}, A_{2}, A_{3} are mutually exclusive and exhaustive events.

P(E/A_{1}) = \(\frac{3}{6}=\frac{1}{2}\)

P(E/A_{2}) = \(\frac{1}{4}\)

P(E/A_{3}) = \(\frac{3}{12}=\frac{1}{4}\)

P(A_{2}/E) = P(A_{2}) . P(E/A_{2}) P(A_{1}) . P(E/A_{1}) + P(A_{2}) . P(E/A_{2}) + P(A_{3}) P(E/A_{3})

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