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Problem will be solved if at least one student can solve. 

∴ Probability of solving by one student = 1/2

= 1 – 1/2 = 1/2

Probability of not solving by any student = 1 – 1/3 = 2/3

Similarly, probability of not solving by third student = 1 – 1/4 = 3/4

Probability of not solving by any one of them = 1/2 x 2/3 x 3/4 = 1/4

∴ Probability of solving by at least one of them = 1 – 1/4 = 3/4

As , S = A ∪ B 

∴ P(A ∪ B) = P(S) = 1 

As A and B are mutually exclusive events. 

∴ P(A ∪ B) = P(A) + P(B) 

⇒ 1 = P(A) + 3P(A) [∵ P(A) = 1/3 P(B)] 

4P(A) = 1 

∴ P(A) = 1/4 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice

3P(A) = 2P(B) = P(C) = k(say) 

∴ P(A) = \(\frac{k}{3}\)

P(B) = \(\frac{k}{2}\)

And P(C) = k 

As events A,B and C are mutually exclusive and exhaustive, 

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1 

⇒ 1 = \(\frac{k}{3} + \frac{k}{2} + {k}\)

⇒ 1 = \(\frac{2k+3k+6k}{6}\) = \(\frac{11k}{6}\) 

⇒ k = \(\frac{6}{11}\) 

As, P(A) =\( \frac{k}{3}\) = \(\frac{1}{3}\times\frac{6}{11}\) = \(\frac{2}{11}\)

Our answer matches with option (b) 

∴ Option (b) is the only correct choice 

The bag contains 5 red and 4 black balls, i.e., 5 + 4 = 9 balls. 

(i) 2 balls can be drawn from 9 balls with replacement in ⁹C₁ X ⁹C₁ ways.

∴ n(S) = ⁹C₁ X ⁹C₁ = 9 × 9 = 81

Let event A: Balls drawn are red. 2 red balls can be drawn from 5 red balls with replacement in ⁵C₁X ⁵C₁ ways.

∴ n(A) = ⁵C₁X ⁵C₁ = 5 × 5 = 25

∴ P(A) = \(\frac {n(A)} {n (S)} = \frac {25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in  ⁹C₁X ⁸C₁ ways.

∴ n(S) = ⁹C₁X ⁸C₁= 9 × 8 = 72

2 red balls can be drawn from 5 red balls without replacement in

⁵C₁X ⁴C₁ ways.

∴ n(B) = ⁵C₁X ⁴C₁= 5 × 4 = 20

 ∴ P(B) = \(\frac {n(B)} {n (S)} = \frac {20}{72} = \frac 5{18}\)

It is given that 5% of men and 0.25% of women have grey hair. 

Therefore, percentage of people with grey hair = (5 + 0.25) % = 5.25%

Probability that the selected haired person is a male =5/5.25=20/21

Let no. of times of tossing a coin be n.

Here, Probability of getting a head in a chance = p = 1/2

Probability of getting no head in a chance = q = 1 - 1/2 = 1/2

Now, P (having at least one head) = P (X ≥ 1)

= 1 - P(X = 0) = 1 - ⁿC₀P⁰p⁰q^{n - 0} = 1 - 1.1.(1/2)ⁿ = 1 - (1/2)ⁿ

From question

1 - (1/2)ⁿ > 80/100

⇒ 1 - (1/2)ⁿ > 8/10 ⇒ 1 - 8/10 > 1/2ⁿ

⇒ 1/5 > 1/2ⁿ ⇒ 2ⁿ > 5 ⇒ n ≥ 3

A man must have to toss a fair coin 3 times.

No. of all possible outcomes = 100

Perfect squares are 4,9,16,25,36, 49,64, 81,100.

No. of favourable outcomes = 9

(i) P(Perfect square) = 9/100

(ii) P(odd number not less than 70) = 16/100 = 4/25 

Let A be Event for getting number on dice

And, B be Event that a spade card is selected

A={2,4,6}

B={13}

Since, P(A)= 3/6 = 1/2

P(B) = 13/52 = 1/4

We know, If E and F are two independent events then, P(AՌB)=P(A).P(B)

P(A ∩ B) =  1/2 x 1/4 = 1/8 

Hence, P(E₁∩ E₂) = 1/8

Total number of outcomes = 52

(i) Let E₄ be the event of getting a queen of black suit.

Number of favorable outcomes = 2

Therefore, P(getting a queen of black suit) = P(E₄) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_4}{number\,of\,all\,possible\,outcomes}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Thus, the probability of getting a queen of black suit is \(\frac{1}{26}\).

(ii) let E₅ be the event of getting a jack of hearts.

Number of favorable outcomes = 1

 Therefore, P(getting a queen of black suit) = P(E₅) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_5}{number\,of\,all\,possible\,outcomes}\) = \(\frac{1}{52}\)

Thus, the probability of getting a jack of heart is \(\frac{1}{52}\).

(iii) let E₆ be the event of getting a spade.

 Number of favorable outcomes = 13

 Therefore, P(getting a queen of black suit) = P(E₆) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_6}{number\,of\,all\,possible\,outcomes}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

Thus, the probability of getting a spade is \(\frac{1}{4}\).

There are total number of batteries , n = 8

Number of dead batteries are = 3

Probability of dead batteries is 3/8

Now, If two batteries are selected without replacement and tested

Then, Probability of second battery without replacement is 2/7

Required probability =  3/8 x 2/7 = 3/28

Let A be the event that the sum of numbers on the dice was less than 6

And, B be the event that the sum of numbers on the dice is 3

A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)

N(A)=10

B={(1,2)(2,1)

n(B)=2

Required probability = nB/nA

Required probability = 2/10

Hence, The probability is  1/5

In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.

Number of cards = 52

Number of queen = 4

Probability of queen out of 52 cards = 4/52

Now, According to the question,

A deck of card shuffled again with replacement, then

Probability of getting queen is , 4/52

Therefore, The probability , that both cards are queen is , [4/52 x 4/52]

Hence, Probability is 1/13 x 1/13.

P(A’) = 2P(B’) = 3P(A ∩ B) = 0.6

∴ P(A’) = 0.6,
2P(B’) = 0.6 .
3P(A ∴ B) = 0.6

∴ P(B’) = \(\frac{0.6}{2}\) = 0.3
P(A ∩ B) =\( \frac{0.6}{3}\) = 0.2

P(A – B) = P(A ∩ B’)
= P(A) – P(A ∩ B)
= [1 -P(A’)] – P(A ∩ B)
= [1 – 0.6] – 0.2
= 0.4 – 0.2
= 0.2

P(B – A) = P(A’ ∩ B)
= P(B) – P(A ∩ B)
= [1 – P(B’)] – P(A ∩ B)
= [1 – 0.3] – 0.2
= 0.7 – 0.2
= 0.5

Correct option is (b) 0

Correct option is (d) Impossible event

Answer: (c) 0.3

S and T being independent events, 

P(S ∩ T) = P(S) × P(T) = 0.3 × 0.4 

P(S/T) = \(\frac{P(S\,\cap\,T)}{P(T)} = \frac{0.3\times 0.4}{0.4}\)  = 0.3

Number of students having blood group AB =

3 Total number of students = 30

Hence, required probability, P = 3/30 = 1/10

Total ways to kept 3 letters in 3 envelops

∴ Probability of kept the letter in right way = 1/31

= 1/(3 x 2) = 1/6

Thus Prob. of kept the letter in wrong way 

= 1 – 1/6 = 5/6

Total no. of integers = 100

∴ n (S) = 100

Favourable cases = (Divisible by 6) + (Divisible by 8) – (Deivisible by 24) 

= {6, 12, 18, …, 198} + {8, 16, 24, …,200} – {24, 48, …, 192}

by formula l = a(n – 1 )d

= 33 + 25 – 8 = 50

∴ Required Prob. = 50/200 = 1/4

Answer is (D) 4/5

Required probability

= ⁹C₂/¹⁰C₂ = 36/45

= 4/5

By arranging five letters of word ANGLE

= 5 ! = 5 × 4 × 3 × 2 = 120

All vowels ocurs together,then

Fovourable cases = 4 ! × 2 !

= 4 × 3 × 2 × 2 = 48

∴ Required Probability = 48/120 = 2/5

(i) Total no of possible outcomes = 6 {1, 2, 3, 4, 5, 6}

E ⟶ Event of getting a prime no.

No. of favorable outcomes = 3 {2, 3, 5}

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

P(E) = 3/6 = 1/2

(ii) E ⟶ Event of getting 2 or 4.

No. of favorable outcomes = 2 {2, 4}

Total no. of possible outcomes = 6

Then, P(E) = 2/6 = 1/3

(iii) E ⟶ Event of getting a multiple of 2 or 3

No. of favorable outcomes = 4 {2, 3, 4, 6}

Total no. of possible outcomes = 6

Then, P(E) = 4/6 = 2/3

(iv) E ⟶ Event of getting an even prime no.

No. of favorable outcomes = 1 {2}

Total no. of possible outcomes = 6 {1, 2, 3, 4, 5, 6}

P(E) = 1/6

(v) E ⟶ Event of getting a no. greater than 5.

No. of favorable outcomes = 1 {6}

Total no. of possible outcomes = 6

P(E) = 1/6

(vi) E ⟶ Event of getting a no. lying between 2 and 6.

No. of favorable outcomes = 3 {3, 4, 5}

Total no. of possible outcomes = 6

P(E) = 3/6 = 1/2

Sample space, S = 6 

(i) Number of event of getting a prime number, n(E) = 3

∴ p = \(\frac{n(E)}{n(S)} \) = \(\frac{3}{6} \) = \(\frac{1}{2}\)

(ii) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{2}{6} \) = \(\frac{1}3\)

(iii) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{4}{6} \) = \(\frac{2}3\)

(iv) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{1}{6} \)

(v) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{1}{6} \)

(vi) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{3}{6} \) = \(\frac{1}{2}\)

As total number of cards = 16 

2 cards can be drawn in ¹⁶C₂ ways = 120 ways

As we need to find the probability for the event E such that at least one of them is an ace. 

We can solve this problem using negation. 

We will find the probability for event that both cards are ace (E’) 

Automatically 1 – P(E’) will give P(E) 

For finding P(E’) we will select both cards out of 4 ace cards 

∴ P(E’) = \(\frac{^4C_2}{120}= \frac{6}{120}=\frac{1}{20}\) 

∴ P(E) = 1 – \(\frac{1}{20}\) = \(\frac{19}{20}\) 

Hence, 

P(E) = \(\frac{19}{20}\) 

As our answer matches only with option (c) 

∴ Option (c) is the only correct choice. 

As 3 dice are thrown so there are 6×6×6 = 216 possibilities. 

Let E denote the event of getting a total score of 5. 

{(1,1,3),(1,2,2)} 

As (1,1,3) can be arranged in \(\frac{3!}{2!}\) ways = 3

and (1,2,2) can be arranged in   \(\frac{3!}{2!}\) ways = 3

∴ n(E) = 6 

Hence, 

P(E) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.