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We have,

P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6

We know that,

P(B|A) × P(A) = P(B ∩ A)

⇒ 0.6 × 0.4 = P(B ∩ A)

⇒ P(B ∩ A) = 0.24

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

= 0.4 + 0.8 – 0.24

= 0.96

Hence, the correct option is D.

Given that P (A) = 0.35 and P (B) = 0.45

∵ The events A and B are mutually exclusive then P (A ⋂ B) = 0

(a) To find (a) P (A′)

We know that,

P (A) + P (A’) = 1

⇒ 0.35 + P(A’) = 1 [given]

⇒ P (A’) = 1 – 0.35

⇒ P (A’) = 0.65

(b) To find (b) P (B′)

We know that,

P (B) + P (B’) = 1

⇒ 0.45 + P (B’) = 1

⇒ P (B’) = 1 – 0.45

⇒ P (B’) = 0.55

(c) To find (c) P (A ⋃ B)

We know that,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ P (A ⋃ B) = 0.35 + 0.45 – 0 [given]

⇒ P (A ⋃ B) = 0.80

(d) To find (d) P (A ⋂ B)

It is given that A and B are mutually exclusive events.

∴ P (A ⋂ B) = 0

(e) To find (e) P (A ⋂ B’)

P (A ⋂ B’) = P (A) – P (A ⋂ B)

= 0.35 – 0

= 0.35

(f) To find (f) P (A’ ⋂ B’)

P (A’ ⋂ B’) = P (A ⋃ B)’

= 1 – P (A ⋃ B)

= 1 – 0.8 [from part (c)]

= 0.2

Given A and B are two mutually exclusive events 

And, P(A) = 0.35 P(B) = 0.45 

By definition of mutually exclusive events we know that: 

P(A ∪ B) = P(A) + P(B) 

We have to find

i) P(A ∪ B) = P(A) + P(B) = 0.35 + 0.45 = 0.8 

ii) P(A ∩ B) = 0 {∵ nothing is common between A and B} 

iii) P(A ∩ B’) = This indicates only the part which is common with A and not B ⇒ This indicates only A. 

P(only A) = P(A) – P(A ∩ B) 

As A and B are mutually exclusive So they don’t have any common parts 

⇒ P(A ∩ B) = 0 

∴ P(A ∩ B’) = P(A) = 0.35 

iv) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law} 

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Number of possible outcomes, n(S) = 2 

Probability of getting a head, P(E’) = \(\frac{1}{2}\) = probability of getting a tail

Since both probabilities are equal for both of the events, therefore both of the players have equal chances to win the game and thus this is a fair way to make this decision that who will play first.

Number of balls in the bag = 3 + 5 = 8

P(that the drown ball is not red) = 5/8

It is given that

Number of times boundary was hit = 6

Total number of balls played = 30

So we get

Number of times boundary was not hit = 30 – 6 = 24

We know that

Probability that the batsman did not hit a boundary = number of times boundary was not hit/ total number of balls played

By substituting the values

Probability that the batsman did not hit a boundary = 24/30

By division

Probability that the batsman did not hit a boundary = 4/5 = 0.8

Total numbers of cards = 18 [odd numbers between 1 and 36]

(i) Favorable numbers = prime number less than 15 = 3, 5, 7, 11, 13

Numbers of favorable outcome = 5

P(prime number less than 15) = 5/18

(ii) Numbers divisible by 3 and 5 = 15 and 30

Since given set contains only odd numbers so exclude 30.

Numbers of favorable outcome = 1

P (getting a number divisible by 3 and 5) = 1/18

Answer is (B)

This cannot be the probability of an event because possibility of event should not be less than 0 and more than 1. Hence -1.5 is lesser than 0.

A box contains 5 red marbles, 8 white marbles and 4 green marbles.

∴ Total number of marbles, n(S) 

= 5 + 8 + 4 = 17 

(i) Probability that the 1 red marble drawn is n(A) = 5 

∴ Probability, P(A) = \(\frac{n(A)}{n(S)} = \frac{5}{17}\)

(ii) Possibility that 1 white marble drawn, n(B) = 8 

∴ Probability, P(B) = \(\frac{n(B)}{n(S)} = \frac{8}{17}\)

(iii) Possibility that 1 not green marble ? 

P(C) = 17 – 4 = 13 (∵ Except 4 green marbles) 

∴ Probability, P(C) = \(\frac{n(c)}{n(S)} = \frac{13}{17}\)

If P(E) = 0.05, then P(bar E)=? 

But, P(E) + P(bar E) = 1 . 

∴ P(bar E) = 1 – 0.05 

∴ P(bar E) = 0.95.

(i) Bag has only lemon flavoured candies. It has no orange candies. 

∴ Possibility, P(E) = 0. 

(ii) A lemon flavoured candy is possible. Because Bag contains all lemon flavoured candies. 

∴ Possibility, P(F) = 1.

A bag contains 3 red balls, 5 black balls. 

Totally there are 8 balls. 

∴ n(S) = 8 

i) Possibility that the red ball drawn, n(E) = 3 

∴ Probability, P(E) = \(\frac{n(E)}{n(S)} = \frac{3}{8}\)

(ii) Possibility that the 1 black ball drwn is n(F) = 5 

∴ Probability, P(F) = \(\frac{n(F)}{n(S)} = \frac{5}{8}\)

Let E be the event of 2 students having the same birthday and _E’ be the event of not getting the same birthday 

Given: 

P(E) = 0.992 

We know that, 

P(E) + P(E’) = 1 

∴ P(E’) = 1 – P(E) 

= 1 – 0.992 

= 0.008

No. of students in the group is 3. 

Probability of 2 students not having the same birthday is 0.992. 

Then possibility of having the same birthday = 1 – 0.992 = 0.008.

It is a case of Bernoulli trials, where success is crossing a hurdle successfully without knocking it down and n = 10.

p = P(success) = 5/6

q = 1 - p = 1 - 5/6 = 1/6

Let X be the random variable that represents the number of times the player will knock down the hurdle. 

Clearly, X has a binomial distribution with n = 10 and p = 5/6

Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n

P(X = r) = ¹⁰C_r(1/6)^r(5/6)^10-r

P (player knocking down less than 2 hurdles) = P( x < 2) 

= P(0) + P(1) = ¹⁰C₀p⁰q¹⁰ + ¹⁰C₁p¹q⁹

= (5/6)¹⁰ + 10(1/6)¹(5/6)⁹ = (5/6)⁹[5/6 + 10/6] = (5/6)⁹ x 15/6 = 5/2 x (5/6)⁹ = 5¹⁰/(2 x 69)

Correct answer is D.

Total penalty kicks = 10

Total goals = 4

Let E be the event of converting a penalty kick into a goal = \(\frac{4}{10}\) = \(\frac{2}{5}\)

Total no. of cards in the deck = 52

Number of red cards = 26

No. of cards drawn = 2 simultaneously

∴ X = value of random variable = 0, 1, 2

Mean = μ = ∑x_iP(X) = 1

Variance = σ² = ∑x²_iP(X) - μ²

= 152/102 - 1 = 50/102 = 25/51 = 0.49

No. of prime numbers between 20 and 

30 = 23, 29 

Total numbers = 10

Probability = 2/10 = 1/5

Correct option is: A) 1/10

Answer: (D) = \(\frac{55}{181}\)

Let E : Event of selecting red and one white ball 

Probability of selecting a box = P(B₁) = P(B₂) = P(B₂) = \(\frac{1}{3}\)

Probability of selecting 1 Red and 1 White ball from box B₁

_{= \(P\big(\frac{E}{B_1}\big) = \frac{^1C_1 \times ^3C_1}{^6C_2} = \frac{1 \times 3 \times 2}{6 \times 5} = \frac{1}{5}\)}  

_{= \(P\big(\frac{E}{B_2}\big) = \frac{^2C_1 \times ^3C_1}{^9C_2} = \frac{2 \times 3 \times 3}{9 \times 8} = \frac{1}{6}\)}  

= \(P\big(\frac{E}{B_3}\big) = \frac{^3C_1 \times ^4C_1}{^{12}C_2} = \frac{3 \times 4 \times 2}{12 \times 11} = \frac{2}{11}\)

\(\therefore\)  \(P\big(\frac{B_2}{E}\big) = \frac {P(B_2) \times P\big(\frac{E}{B_2}\big)}{P(B_1) \times P\big(\frac{E}{B_1}\big) +P(B_2) \times P\big(\frac{E}{B_2}\big) +P(B_3) \times P\big(\frac{E}{B_3}\big)}\)   (Using Bayes Theorem)

= \(\frac{\frac{1}{3} \times \frac{1}{6}}{\frac{1}{3} \times \frac{1}{5} +\frac{1}{3} \times \frac{1}{6} +\frac{1}{3} \times \frac{2}{11}}\)  = \(\frac {\frac{1}{6}}{\frac{66+55+30}{330}}\) = \(\frac{\frac{1}{6}}{\frac{181}{330}}\)

= \(\frac{55}{181}\)

The letters of the word EQUATION can be arranged in 8! ways. 

∴ n(S) = 8! 

There are 5 vowels and 3 consonants. 

(i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N 

Let us consider all vowels as one unit. 

So, there are 4 units, which can be arranged in 4! ways. 

Also, 5 vowels can be arranged among themselves in 5! ways. 

∴ n(A) = 4! × 5! 

Required probability = P(A)

\(\frac {n(A)}{n(S)} = \frac {4!\times5!} {8!} = \frac {1}{14}\)

(ii) B: arrangement start with a vowel and ends with a consonant. 

First and last places can be filled in 5 and 3 ways respectively. 

Remaining 6 letters are arranged in 6! Ways. 

∴ n(B) = 5 × 3 × 6! 

Required probability = P(B)

\(\frac {n(B)}{n(S)} = \frac {5\times3\times6!} {8!} = \frac {15}{56}\)