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Four cards are drawn at random from a pack of 52 cards. What is the probability of getting all four cards of the same suit?

A.

2860/27725

B.

2560/270725

C.

2360/270725

D.

2860/270725

Answer with explanation

Answer: Option DExplanation

Total No. Of cards =52

No.Of cards for each suit = 13

Total types Of French suit = 4(hearts,spades, diamonds,clubs)

Prob. = Conditional case /total case

For this conditional case of selecting all four same suit,it can be either hearts or spades or clubs or diamonds

So Total no. Of conditional case= no. of case of selecting all four cards of hearts suit +no.of case of selecting all four cards of spades suit +no.of case of selecting all four cards of clubs suit + no. of case of selecting all four cards of diamonds suit.

= 13C4+13C4+13C4+ 13C4

=4*(13C4)=2860

Total no. of case= 52C4=270725

So, probability of getting all four cards of the same suit=2860/270725

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In a charity show tickets numbered consecutively from 101 through 350 are placed in a box.

What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A.

0.285

B.

0.40

C.

100/249

D.

99/250

Answer with explanation

Answer: Option BExplanation

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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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