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A speaks truth in 75% of cases and BB in 80% of cases. In what percent of cases are they likely to contradict each other in narrating the same event?

A.

35%

B.

5%

C.

15%

D.

65%

Answer with explanation

Answer: Option AExplanation

Different possible cases of contradiction,

AA speaks truth and BB does not speaks truth.

Or, AA does not speak truth and BB speaks truth.

=(3/4×1/5)+(1/4×4/5)=(3/4×1/5)+(1/4×4/5)

=3/20+4/20=3/20+4/20

=7/20=7/20

=35%

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In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

A.

0.64

B.

0.68

C.

0.65

D.

0.67

Answer with explanation

Answer: Option DExplanation

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67

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Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

A.

5/13

B.

7/15

C.

7/12

D.

7/13

Answer with explanation

Answer: Option DExplanation

We need to find out P(B or 6)

Probability of selecting a black card = 26/52

Probability of selecting a 6 = 4/52

Probability of selecting both a black card and a 6 = 2/52

P(B or 6) = P(B) + P(6) – P(B and 6)

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13.

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Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.

A.

6/ 26

B.

3/ 26

C.

4/ 26

D.

4/ 15

Answer with explanation

Answer: Option DExplanation

Let E1, E2, E3 and A are the events defined as follows.

E1 = First bag is chosen

E2 = Second bag is chosen

E3 = Third bag is chosen

A = Ball drawn is red

Since there are three bags and one of the bags is chosen at random, so P (E1) = P(E2) = P(E3) = 1 / 3

If E1 has already occurred, then first bag has been chosen which contains 3 red and 7 black balls. The probability of drawing 1 red ball from it is 3/10. So, P (A/E_{1}) = 3/10, similarly P(A/E_{2}) = 8/10, and P(A/E_{3}) = 4/10. We are required to find P(E_{3}/A) i.e. given that the ball drawn is red, what is the probability that the ball is drawn from the third bag by Baye’s rule

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** **A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the problem is solved?

A.

112/147

B.

162/147

C.

132/147

D.

122/147

Answer with explanation

Answer: Option DExplanation

Probability of the problem getting solved = 1 – (Probability of none of them solving the problem)

Probability of problem getting solved = 1 – (5/7) x (3/7) x (5/9) = (122/147)

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In a town 45% people speak English, 30% speak Hindi and 20% speak both English and Hindi. One person is selected at random. Find the probability that he speaks English, if it is known that he speaks Hindi.

A.

1/3

B.

2/3

C.

5/6

D.

2/5

Answer with explanation

Answer: Option BExplanation

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There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?

A.

191/1547

B.

180/1547

C.

280/1547

D.

189/1547

Answer with explanation

Answer: Option CExplanation

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When 4 fair coins are tossed together what is the probability of getting at least 3 heads?

A.

1/4

B.

3/4

C.

5/16

D.

3/8

Answer with explanation

Answer: Option CExplanation

When 4 fair coins are tossed simultaneously, the total number of outcomes is 2^{4} = 16

At least 3 heads implies that one can get either 3 heads or 4 heads.

One can get 3 heads in ^{4}C_{3} = 4 ways and can get 4 heads in ^{4}C_{4} = 1 ways.

∴ Total number of favorable outcomes = 4 + 1 = 5

====>>> The required probability = 1/4.

or

The answer is 5/16 because in total there are 16 possibilities when tossing a coin 4 times. I.e. 2⋅2⋅2⋅2=242⋅2⋅2⋅2=24.

And the possibilities of getting atleast three head is 5 that is we can get HHHT, HHTH, HTHH, THHH, and HHHH

Thus, the answer is 5/16.

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A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A.

158/435

B.

316/435

C.

17/87

D.

136/345

Answer with explanation

Answer: Option BExplanation

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Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

A.

1/4

B.

1/2

C.

3/4

D.

7/4

Answer with explanation

Answer: Option CExplanation

Total number of cases = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So Probability = 27/36 = 3/4

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Two unbiased coins are tossed. What is the probability of getting at most one tail?

A.

3/4

B.

1/2

C.

3/2

D.

1/4

Answer with explanation

Answer: Option AExplanation

Total 4 cases = [HH, TT, TH, HT]

Favourable cases = [HH, TH, HT]

Please note we need atmost one tail, not atleast one tail.

So probability = 3/4

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A pot contains 5 white and 3 red balls while another pot contains 4 white and 6 red balls. One pot is chosen at random and a ball is drawn from it. If the ball is white, what is the probability that it is from the first pot?

A.

1/5

B.

5/16

C.

25/41

D.

41/80

Answer with explanation

Answer: Option CExplanation

The probability of choosing one pot = 1/2

The probability of choosing white ball in first pot = 1/2 × [ 5C1 / 8C1] = 5/16

The probability of choosing white ball

= 1/2 × [5C1 / 8C1] + 1/2 × [4C1 / 10C1]

= 1/2 × [5/8] + 1/2 × [4/10]

= 5/16 + 1/5 = 41/80 ∴ The probability that it is from the first pot = (5/16) / (41/80)

= 5/16 × 80/41 = 25/41

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In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?

A.

16/21

B.

9/14

C.

5/21

D.

5/14

Answer with explanation

Answer: Option CExplanation

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅ ==>> (10 * 6)/252 = 5/21

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A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is (1/7) and the probability of wife’s selection is (1/5). What is the probability that only one of them is selected ?

A.

3/4

B.

2/7

C.

4/5

D.

1/7

Answer with explanation

Answer: Option BExplanation

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A dice is thrown, what is the probability that the number obtained is a prime number.

A.

1/3

B.

1/6

C.

1/8

D.

1/2

Answer with explanation

Answer: Option DExplanation

Dice is thrown, the total possible outcomes = 6.

Favorable outcomes = 3 i.e. (2,3,5). Probability = 3 / 6 = 1 / 2

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Let S be the sample space and two mutually exclusive events A and B be such that A U B = S. If P(.) denotes the probability of the event. The maximum value of P(A)P(B) is ______

A.

0.5

B.

0.25

C.

0.125

D.

0.225

Answer with explanation

Answer: Option BExplanation

Sample Space(S) – A set of all possible outcomes/events of a random experiment.

Mutually Exclusive Events – Those events which can’t occur simultaneously.

P(A)+P(B)+P(A∩B)=1 Since the events are mutually exclusive, P(A∩B)=0.

Therefore, P(A)+P(B)=1

Now, we know that AM >= GM So, (P(A)+P(B))/2 >= sqrt(P(A)*P(B)) P(A)*P(B) <= 1/4

Hence max(P(A)*P(B)) = 1/4.

We can think of this problem as flipping a coin, it has two mutually exclusive events ( head and tail , as both can’t occur simultaneously).

And sample space S = { head, tail } Now, let’s say event A and B are getting a “head” and “tail” respectively.

Hence, S = A U B. Therefore, P(A) = 1/2 and P(B) = 1/2.

P(A).P(B) = 1 /4 = 0.25.

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An integer is chosen at random from the first fifty integers. What is the probability that the integer chosen is a prime or multiple of 4?

A.

27/50

B.

3/4

C.

11/25

D.

14/35

Answer with explanation

Answer: Option AExplanation

There are 15 prime numbers in the first 50 integers i.e. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

There are 12 integers are multiples of 4 i.e 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44 an 48

∴ required probability= 15/50 + 12/50 = 27/50

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What is the probability of getting a sum 9 from two throws of dice?

A.

2/3

B.

1/3

C.

1/6

D.

1/9

Answer with explanation

Answer: Option DExplanation

Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

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A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident

A.

45%

B.

40%

C.

35%

D.

30%

Answer with explanation

Answer: Option CExplanation

Let A = Event that A speaks the truth

B = Event that B speaks the truth

Then P(A) = 75/100 = 3/4

P(B) = 80/100 = 4/5

P(A-lie) = 1-3/4 = 1/4

P(B-lie) = 1-4/5 = 1/5

Now

A and B contradict each other =

[A lies and B true] or [B true and B lies]

= P(A).P(B-lie) + P(A-lie).P(B)

[Please note that we are adding at the place of OR]

= (3/5*1/5) + (1/4*4/5) = 7/20

= (7/20 * 100) % = 35%

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A.

4/25

B.

8/25

C.

2/25

D.

1/25

Answer with explanation

Answer: Option CExplanation

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In a throw of dice, what is the probability of getting number greater than 5

A.

1/5

B.

1/2

C.

1/6

D.

1/3

Answer with explanation

Answer: Option CExplanation

Number greater than 5 is 6, so only 1 number

Total cases of dice = [1,2,3,4,5,6]

So probability = 1/6

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If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is –

A.

4/13

B.

5/13

C.

19/52

D.

17/52

Answer with explanation

Answer: Option AExplanation

P(SᴜK) = P(S) + P(K) – P(S∩K), where S denotes spade and K denotes king.

P(SᴜK) = 13/52 + 4/52 – 1/52 = 4/13

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