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# Probability Questions

Home > Quantitative Aptitude > Probability > General Questions
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A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spade. Find the probability of the lost card being a spade.

A10/50

B10/53

C11/50

D11/53

Explanation:

Solution: Total cards:52

Now, actual present total cards=total cards-drawn cards=52-2=50

So ,probability=11/50

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A 3*3 grid is coloured using red and blue colours,such that if we rotate the grid about its centre plane by 180 degrees,the grid looks the same.The number of ways to colour the grid this way is:

A256

B64

C16

D32

Explanation:

Here is no explanation for this answer

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There are three boxes containing red, white and mixed balls, which are mislabelled. It is known that one of the boxes contains only white and one only red balls. The third contains a mixture of red and white balls. You are requested to correctly label the boxes with the labels Red, White and Red & White, by picking a sample of one ball from only one box. What is the label on the box you should sample ?

AWhite

BRed

CRed & White

DNot possible to determine from a sample of one ball

Explanation:

With the given information, it is not possible to decide which sample should be chosen.

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Two decks of cards are there. Each deck contains 20 cards, with numbers from 1 to 20 written on them. A card is drawn of random from each deck, getting the numbers x and y What is the probability that log x + log y is a positive integer. Logs are taken to the base 10.

A3/200

B29/200

C29/200

D1/50

Explanation:

Log x + log y = log(xy)
log xy is integer when (x,y) = (1, 10), (10, 1), (10, 10), (5, 20), (20, 5), (2, 5), (5, 2)
So required probability = 7/400

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Four people each roll a four die once. Find the probability that at least two people will roll the same number?

A5/18

B13/18

C1295/1296

DNone of these

Explanation:

The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 * 5 * 4 * 3

Now total possibilities of rolling a dice = 64

The probability that a no one gets the same number = 6*5*4*364 = 518
So, the probability that at least two people gets same number = (1?518) = 1318

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A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?

A291/1100

B292/1100

C290/1100

D301/1100

Explanation:

Numbers which dont have 2 from 1 to 9 = 8 Numbers which dont have 2 from 10 to 99:

Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbres.

Numbers which dont have 2 from 100 to 999 =_ _ _ = 8 * 9 * 9 = 648 Numbers which dont have 2 from 1000 to 1099 =10_ _ = 9 * 9 = 81 Finally 1100 does not have 2. So 1.

Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810 Tickets with 2 in them = 1100 - 810 = 290

Required probability = 290 / 1100

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Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle?

A440/455

B412/455

C449/455

D438/455

Explanation:

Total ways of selecting 3 dots out of 15 is 15C3 = 455 If 3 dots are collinear then triangle may not be formed. Now look at the above diagram.
If we select any 3 dots from the red lines they may not form a triangle. They are 3 x 5C3 = 30.
If we select the three letters from blue lines, they may not form a triangle.
They are in total 5 ways. Also there are 6 others lines which don't form a triangle. Also another two orange lines.
Total = 30 + 5 + 6 + 2 = 43. So we can form a triangle in 455 - 43 = 412.

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In this question, A^B refers to A raised to the power B.Ten tickets numbered 1, 2, 3, ..., 10. Six tickets are selected at random one of a time with replacement. The probability of the largest number appearing on the selected ticket is 7 is

A(7^6 +1)/10^6

B(7^6 - 6^6)/10^6

C(7^6+6^6)/10^6

D6^6/10^6

Explanation:

Let's first find out probability of that maximum number being any number between 1 to 7.
P(1 to 7) = (7/10) * (7/10) * (7/10) ... 6 times

Now find out probability of that maximum number being any number between 1 to 6.
P(1 to 6) = (6/10) * (6/10) * (6/10) ... 6 times

Now, probability that maximum number is exactly 7
= P(1 to 7) - P (1 to 6)
= (7^6 - 6^6) / 10^6

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The probability that a number selected at random from first 50 natural numbers is a composite number is

A21/25

B17/25

C4/25

D8/25

Explanation:

The number of exhaustive events = 50C1 = 50.
We have 15 primes from 1 to 50.
Number of favourable cases are 34.
Therefore, Required probability = 34/50 = 17/25.

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A factory produces 1 defective bulb out of 10 bulbs a yr. if it produces 870 bulbs a year, how many defective bulbs are produced?

A78

B90

C87

D80

Explanation:

10 out of 1 is defective
20 out of 2 is defective
100 out of 10 is defective
and so on
800 out of 80 is defective
70 out of 7 is defective
80+7=87

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