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Total outcomes = 6

Prime numbers = 2,3,5 = 3

P(Prime no.) = 3/6 = 1/2

Correct option is (D) 1 – x

P(A) = x

\(\because\) P(A) + P(A') = 1

\(\Rightarrow\) P(A') = 1 - P(A)

= 1 - x

Correct option is  D) 1 – x

P(winning the game) = 0.08

P(losing the game ) = 1 - 0.08 

= 0.92

Correct option is (B) 1/2

There are only two possibilities either you win prize or you are not win the prize in any contest.

\(\therefore\) The probability of winning a prize \(=\frac{1}{2}.\)

Correct option is  B) 1/2 

Correct option is: B) 1/2

P(E) + P(not E) = 1 

0.59 + P(not E) = 1 

P(not E) = 1 – 0.59 = 0.41

Correct option is: B) \(\frac {11}{15}\)

P (A) + P (\(\overline A\)) = 1

\(\Rightarrow\) P (\(\overline A\))  = 1 - P(A)

= 1 - \(\frac 4{15}\) (\(\because\) P (A) = \(\frac 4{15}\))

= \(\frac {11}{15}\)

Correct option is: B) \(\frac{11}{15}\)

Let event A: The girl cracks the National Level exam. 

∴ P(A) = 0.42 

Let event B: The girl cracks the State Level exam. 

∴ P(B) = 0.54 Also, P(A ∩ B) = 0.11 

(i) P(the girl cracks at least one of the two exams)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

= 0.42 + 0.54 – 0.11 

= 0.85

(ii) P(the girl cracks only one of the two exams) 

= P(A) – P(B) – 2P(A ∩ B) 

= 0.42 + 0.54 – 2(0.11) 

= 0.74 

(iii) P(the girl cracks none of the exams) 

= P(A’ ∩ B’) 

= P(A ∪ B)’ 

= 1 – P(A ∪ B) 

= 1 – 0.85 

= 0.15

An experiment succeeds thrice as often as it fails. 

⇒ p = P(getting success) = and 

q = P(getting failure) = Here, number of trials = n = 5 

By binomial distribution, we have

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

P(X = r) = ⁵C_r(3/4)^r(1/4)^{5 - r}

Now , P(getting at least 3 success) = P(X = 3) + P(X = 4) + P(X = 5)

= ⁵C₃(3/4)³(1/4)² + ⁵C₄(3/4)⁴(1/4)¹ + ⁵C₅(3/4)⁵

= (3/4)³[10 x 1/16 + 15 x 1/16 + 9 x 1/16] = 27/64 x 34/16 = 459/512

We know that

P(A') = 1 - P(A)
Given , 

\({P(A) \over P(A')} = {4 \over 5}\)

⇒ \({P(A) \over 1-P(A)} = {4 \over 5}\)

⇒\(5P(A) = 4 - 4P(A)\)

⇒\(9P(A) = 4\)

⇒\(P(A) = {4 \over 9}\)

(d) \(\frac{1}{132}\)

Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! 

If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the remaining, i.e., 6 girls can be arranged in a row in 6! ways. There are 5 places between the 6 girls and 2 on the extreme ends, where the entity of 6 boys can be placed, i.e., for the single entity of 6 boys, we have 7 places where they can be arranged in 7 ways and also amongst themselves they can be arranged in 6! ways. 

∴ No. of ways of arranging 6 boys and 6 girls in a row where the 6 boys are together = 6! × 7 × 6! 

∴ Required probability = \(\frac{6!\times7\times6!}{12!}\) = \(\frac{7\times6\times5\times4\times3\times2}{12\times11\times10\times9\times8\times7}\) = \(\frac{1}{132}\).

A and B are independent events.

∴ P(A ∩ B) = P(A) – P(B)

∴ 0.12 = P(A) × 0.3

∴ P(A) = \(\frac{0.12}{0.3}\) = 0.4

Now, according to the law of addition of probability,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.4 + 0.3 – 0.12

= 0.58

Total cards = 49

Odd No. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 = 25

(i) P(odd number) = 25/49

Multiple of 5 are 5, 10 ,15, 20, 25, 30, 35, 40, 45 = 9

(ii) P(multiple of 5) = 9/49

Only even prime number is = 2

(iii) P(even prime) = 1/49

Sample space S = GG, GB, BG, BB (optional)

P(atleast one girl) = 3/4 

366 days = 52 weeks + 2 days

2 days can be MT, TW, WTh,ThF, FS, SS, SM = 7

=> P = 2/7

Answer is (B)

P(A’B’) = P(A’∩B’) 

= PA ∪ B

= 1 – P (A∪B) 

= 1 – P(A) – P(B) – P(A) – P(B) 

= 1 – P (A) – P (B) [1 – P (A)]

= [1 – P (A)] [1 -P(B)]

Answer is (D)

Even Prime no is (2,2) 

P(even prime no) = \(\frac{1}{36}\)

As box contains 26 balls. Let S represents the sample space. 

∴ n(S) = 26 

Let W denotes the event of drawing a white ball and R represents event of drawing red ball. 

As, n(W) = 10 and n(R) = 6 

∴ P(W) = \(\frac{10}{26}\) = \(\frac{5}{13}\) 

And P(R) = \(\frac{6}{26}\) = \(\frac{3}{13}\)  

As both events have nothing in common so we can say that they are mutually exclusive. 

We need to find the probability of the event such that ball drawn is red or white. 

P(Red or White) = P(R∪W) 

As events are mutually exclusive 

∴ P(R ∪ W) = P(R) + P(W) 

⇒ P(R∪W) =   \(\frac{3}{13}\)  +   \(\frac{5}{13}\)  = \(\frac{8}{13}\) 

Given, odds in favour of A is \(\frac{P(A)}{P(\bar A)}\) = \(\frac{1}{3}\)

⇒ \(\frac{P(A)}{1-P(A)} = \frac{1}{3}\)

 ⇒ 1 – P(A) = 3P(A) 

⇒ 4P(A) = 1 

⇒ P(A) = \(\frac{1}{4}\)

Odds in favour of horse B is  \(\frac{P(B)}{P(\bar B)}\) = \(\frac{1}{4}\) 

\(\frac{P(B)}{1-P(B)} = \frac{1}{4}\) 

⇒ 1 – P(B) = 4P(B) 

⇒ 5P(B) = 1 

⇒ P(B) = \(\frac{1}{5}\) 

Odds in favour of horse C is \(\frac{P(C)}{P(\bar C)}\) = \(\frac{1}{5}\) 

⇒ \(\frac{P(C)}{1-P(C)} = \frac{1}{5}\) 

⇒ 1 – P(C) = 5P(C) 

⇒ 6P(C) = 1 

⇒ P(C) = \(\frac{1}{6}\) 

Odds in favour of horse D is \( \frac{P(D)}{P(\bar D)}\) = \(\frac{1}{6}\) 

⇒ \(\frac{P(D)}{1-P(D)} = \frac{1}{6}\)  

⇒ 1 – P(D) = 6P(D) 

⇒ 7P(D) = 1 

⇒ P(D) = \(\frac{1}{7}\) 

We have to find the probability that one of the horses win the race. 

∵ only one horse can win the race ⇒ A ,B,C and D are mutually exclusive events. 

We need to find P(A∪B∪C∪D). 

∵ A ,B,C and D are mutually exclusive events. 

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

= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\) = \(\frac{319}{420}\) 

Hence, 

probability that one of the horses win the race = \(\frac{319}{420}\)

Given that the dice is thrown three times

So, the sample space n(S) = 6³ = 216

Let E₁ be the event when the sum of number on the dice was 6 and E₂ be the event when three 2’s occur.

E₁ = {(1, 1, 4), (1, 2, 3), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)}

n(E₁) =10 and n(E₂) = 1 [Since, E₂ = (2, 2, 2)]

Thus, P(E₂/E₁) = P(E₁ ⋂ E₂)/P(E₁) = (1/216)/(10/216) = 1/10.

Let’s take W₁ and W₂ to be two bags containing (4W, 5B) and (9W, 7B) balls respectively.

Let E₁ be the event that the transferred ball from the bag W₁ to W₂ is white and E₂ the event that the transferred ball is black.

And, E be the event that the ball drawn from the second bag is white.

So, the probabilities are:

P(E/E₁) = 10/17, P(E/E₂) = 9/17

P(E₁) = 4/9 and P(E₂) = 5/9

Now, P(E) = P(E₁).P(E/E₁) + P(E₂).P(E/E₂)

= 4/9 x 10/17 + 5/9 x 9/17 = 40/153 + 45/153 = 85/153 = 5/9

Therefore, the required probability is 5/9.