Explore topic-wise InterviewSolutions in .

We have, sample space S of the experiment is 5 = {(x,y):x,y = l,2,3,4,5,6} 

Then A = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4.1), (4,2), (4,3), (4,4), (4,5), (4,6) (6.1),(6,2),(6,3),(6,4),(6,5),(6,6)} 

B = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (3.1) , (3,2), (3,3), (3,4), (3,5), (3,6), (5.1) , (5,2), (5,3), (5,4), (5,5), (5,6)} 

C = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3) (3.1) , (3,2), (4,1)} 

(i) A’ = B 

(ii) B’ = A 

(iii) A∪B = S 

(iv) A∩B = ϕ 

(v) A – C = {(2,4),(2,5),(2,6),(4,2),(4,3) , (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4),(6,5),(6,6)} 

(vi) B ∪ C = {(1,1), (1,2), ………… (1,6), (2,1), (2,2), (2.3),(3,1), (3,2), (3,3), ……….. (3,6), (4,1), (5,1), (5,2),(5,6)} 

(vii) B ∩ C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}

(viii) Here, C’ = S – C and B’ = A 

∴ A ∩ B’∩ C’ = A ∩ A ∩ C’ 

= A ∩ C’ 

= {(2,4), (2,5), (2,6), (4,2), (4.3), (4,4), …………, (4,6), (6,1), (6,2), ………. , (6,6)}

• When the number of outcomes of the sample space is infinite, it cannot be used to find the probability of an event.
• If the total number of outcomes of the sample space is unknown, the probability of an event cannot be obtained.
• If all possible outcomes of the sample space are not equiprobable, the probability of an event cannot be obtained.

Correct option is (c) P(H ∩ T) = 0.5

Correct option is (a) Mathematical definition

Correct option is (a) P(A ∩ B) = P(B)

Digital store A: 6 LED TV + 4 LCD
TV = 10 TVs are displayed

Digital store A: 6 LED TV + 4 LCD
TV = 10 TVs are displayed

A₁ = Event that a store is selected from the store A and B
∴ p(A₁) = \(\frac{1}{2}\)

A₂ = Event that LCD TV is selected from the store A
∴ p(A₂) = \(\frac{^4C_1}{^{10} C_1} = \frac{4}{10}\)

B₁ = Event that if store A is selected then LCD TV is selected.
∴ p(B₁) = p(A₁) ∙ p(A₂) = \(\frac{1}{2}×\frac{4}{10} = \frac{4}{20}\)

B₂ = Event that if store B is selected then LCD TV is selected
∴ p(B₂) = p(A₁) ∙ p(A₃) = \(\frac{1}{2}×\frac{3}{8} = \frac{3}{16}\)

C1 = Event that if one of the two stores is selected then LCD TV Is selected

= B₁ ∪ B₂

∴ P(C₁) = P(B₁) + P(B₂)

= \(\frac{4}{20}+\frac{3}{16}\)

= \(\frac{16+15}{80}\)

= \(\frac{31}{80}\)

Correct option is (c) 0 ≤ P(A) ≤ 1

One number is randomly selected from the numbers 1 to 100.

∴ Total number of primary outcomes,

n = ¹⁰⁰C₁ = 100

A = Event that the number selected is a single digit number

= {1. 2,3, 4,5,6,78, 9}

∴ m = 9

∴ P(A) =\(\frac{ m}{n} = \frac{9}{100}\)

B = Event that the number selected is a perfect square

= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100)

∴ m = 10

∴ P(B) = \(\frac{ m}{n} = \frac{10}{100}\)

A ∩ B = Event that the selected number is a single digit number and a perfect square

= {1, 4,9}

∴ m = 3

∴ P(A ∩ B) = \(\frac{m}{n}=\frac{3}{100}\)

Now, A ∪ B = Event that the number selected is either a single digit number or a perfect square

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{9}{100}+\frac{10}{100}−\frac{3}{100}\)

= \(\frac{16}{100}=\frac{4}{25}\)

Correct option is (a) P(A ∩ B) = P(A) × P(B)

P(E) = Favourable outcomes/ total no. of possible outcomes

here, favourable outcomes are 5,10,15,20...

So P(multiple of 5) = 4/20 = 1/5 

So, the answer is (c) 1/5.

Correct option is (c) \(\frac{1}{5}\)

Correct option is (d) 6³

Correct option is (b) P(B)

Correct option is (a) Independent events

Correct option is (b) Complementary events

(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1. P(A) + P(B) = 1

(ii) P(A) = 0.46 

Let P(B) be the probability of not happening of event A

We know, 

P(A) + P(B) = 1

P(B) = 1 – P(A) 

P(B) = 1 – 0.46 

P(B) = 0.54 

Hence the probability of not happening of event A is 0.54

Sample space (S) = {1,2, 3, 4, 5, 6, 7, 8} 

∴ n(S) = 8 

i. Let A be the event that the spinning arrow comes to rest at 8.

∴ A = {8}

∴ n(A) = 1

∴ P(A) = \(\frac{n(A)}{n(S)}\)

∴ P(A) = 1/8

ii. Let B be the event that the spinning arrow comes to rest at an odd number.

∴ B = {1,3,5,7}

∴ n(B) = 4

∴ P(B) = \(\frac{n(B)}{n(S)}\) = 4/8

∴ P(B) = 1/2

iii. Let C be the event that the spinning arrow comes to rest at a number greater than 2.

∴ C = {3,4,5,6,7,8}

∴ n(C) = 6

∴ P(C) = \(\frac{n(C)}{n(S)}\) = 6/8

∴ P(C) = 3/4

iv.  Let D be the event that the spinning arrow comes to rest at a number less than 9. 

∴ D = {1,2, 3, 4, 5, 6, 7, 8}

∴ n(D) = 8 

∴ P(D) = \(\frac{n(D)}{n(S)}\) = 8/8

∴ P(D) = 1

∴P(A) = 1/8; P(B) = 1/2; P(C) = 3/4; P(D) = 1

The correct answer is : (C) 1/4

Correct answer is

(C) 1/4

n(S) = 100

Let A be the event that the number chosen is a prime number.

∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

∴ n(A) = 25

∴ P(A) = \(\frac{n(A)}{n(S)} = \frac{25}{100} = \frac{1}{4}\)

Total no. of possible outcomes = 89 {2, 3, 4, …., 90}

(i) Let E ⟶ event of getting a 2 digit no.

No. favourable outcomes = 81 {10, 11, 12, 13, ….., 80}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 81/89

(ii) E ⟶ event of getting a no. which is perfect square

No. favourable outcomes = 8 {4, 9, 16, 25, 36, 49, 64, 81}

P(E) = 8/89

Correct answer is

(A) 1/5

Total number of balls = 18

(i) Let x be number of red balls. So, there are (18 – x) non red balls.

P(getting a non red ball) = (18-x)/18

(ii) if 2 more red balls are out in the bag, then there are 20 total number of balls.

Number of red balls x + 2

So we have,

P(getting a red ball) = 9/8 p(getting a red ball in first option)

Now,

(x+2)/20 = 9/8(x/18)

16x + 32 = 20x

X = 8

Let total number of balls = x

Orange balls in the jar = 10

P(getting an orange ball) = 10/x

Since jar contains only red, blue and orange balls.

P(getting a red ball) + P(getting a blue ball) + P(getting a orange ball) = 1

(Because sum of probability = 1)

1/4 + 1/3 + 10/x = 1

3x + 4x + 120 = 12x

x = 24

Hence there are total 24 balls in the jar.

Total numbers of cards = 100 + 70 + 50 + 30 = 250

(i) Number of favourable outcomes = seventy ₹ 1 coin = 70

P(getting ₹ 1 coin) = 70/250 = 7/25

(ii) Numbers of ₹ 5 coins = thirty ₹ 5 coins = 30

Numbers of favourable outcomes = will not be a ₹ 5 coin = 250 – 30 = 220

P(getting a coin will not be a ₹ 5 coin) = 220/250 = 22/25

(iii) Number of 50-p coins = 100

Number of ₹ 2 coins = 50

Total coins = 100+50 = 150

P (getting 50-p coin or ₹ 2 coins)= 150/250 = 3/5

Correct option is: C) 1

P(E) + P (\(\overline E\)) = P (E) + P (not E) = 1

Correct option is: C) 1

Correct option is: B)\(\frac 5 {36}\)

Since, two dice are thrown.

\(\therefore\) Total possible outcomes = n (s) = 6 \(\times\) 6 = 36.

Outcomes fabourable to event that the sum of numbers appearing on the top of both dices is 8 are 

{(2,6), (3, 5), (4, 4), (5, 3), (6, 2)}

\(\therefore\) Total favourable outcomes = 5.

\(\therefore\) Probability = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac 5 {36}\)

Correct option is: B) \(\frac{5}{36}\)

We have,  \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\) 

Therefore \(P(\cfrac{A}{B})=\cfrac{0.32}{0.5}\)

{multiply numerator and denominator by 2 to convert into the whole number}

= 0.64

We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

From equation, \(P(\cfrac{B}{A})=\cfrac{P(A\cap B)}{P(A)}\)

P (B ∩ A) = \(P(\cfrac{B}{A})\) x P(A) = 0.6 x 0.4

= 0.24

Hence,  \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}=\cfrac{0.24}{0.8}\)

= 0.3

And P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

= 0.4 + 0.8 - 0.24

= 0.96

As, (A ∪ B) = (A ∩ B) 

⇒ A and B are same sets. 

∴ P(A) = P(B) 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice

We have, P(A ⋃ B) = P(A ⋂ B)

By General Addition Rule,

P(A) + P (B) – P(A ⋂ B) = P(A ⋃ B)

⇒ P(A) + P (B) – P(A ⋂ B) = P(A ⋂ B) [given]

⇒ [P(A) – P(A ⋂ B)] + [P(B) – P(A ⋂ B)] = 0

But P(A) – P(A ⋂ B) ≥ 0

and P(B) – P(A ⋂ B) ≥ 0

[∵ P(A ⋂ B) ≤ P(A) or P(B)]

⇒ P(A) – P(A ⋂ B) = 0

and P(B) – P(A ⋂ B) = 0

⇒ P(A) = P(A ⋂ B) …(i)

and P(B) = P(A ⋂ B) …(ii)

From (i) and (ii), we get

∴ P(A) = P(B)

Hence, the correct option is (A).

Given there are 6 boys and 6 girls

\(\therefore\) No of ways in which 6 boys and 6 girls sitting together in a row=7!

6 girls sitting arrangement = 6!

\(\therefore\) required probability = \(\cfrac{7!\times6!}{12!}\)

= \(\cfrac1{132}\)

We have,

P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

⇒ 0.5 = 0.4 + 0.3 – P(A ∩ B)

⇒ P(A ∩ B) = 0.7 – 0.5

⇒ P(A ∩ B) = 0.2

∴ P(B’ ∩ A) = P(B’) P(A)

= [1 – P(B)]× P(A)

[sum of the probabilities of an event and its complement is 1]

= P(A) – P(A)P(B)

= P(A) – P(A ∩ B)

= 0.4 – 0.2

= 0.2

= 1/5

Hence, the correct option is D

Given, S = {1, 2, 3, …, n}

So, P(r ≤ p/s ≤ p) = P(P ⋂ S)/ P(S)

= p – 1/n x n/(n – 1)

= (p – 1)/(n – 1)

Therefore, the required probability is (p – 1)/(n – 1).