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eLitmus Probability questions for eLitmus exam | eLitmus Questions for pH Test

6

A man can hit the target once in four shots. If he fires 4 shots in succession, what is the probability that he will hit target ?

A.

1

B.

1/256

C.

81/256

D.

175/256

See Answer & Explanation Lets Discuss
Correct answer is : D
Explanation

The target is not hit only when he is not able to hit it in any of the four shots. So the probability that he will hit target = 1 - (probability that he will not hit the target in any of the four shots)

= 1 - (3/4)*(3/4)*(3/4)*(3/4)

= 1 - 81/256

= 175/215

 
7

When four dice are thrown, what is the probability that the same number appears on each of them ?

A.

1/36

B.

1/18

C.

1/216

D.

1/5

See Answer & Explanation Lets Discuss
Correct answer is : C
Explanation

When four dice are rolled the total no of outcomes are : 64

All dice show up same number in the following cases: (1,1,1,1), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)

So in above six cases, all dice will show up the same number.

Required probability: 6/64 = 1/216

 
8

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

A.

1/18

B.

64/4032

C.

63/64

D.

1/9

See Answer & Explanation Lets Discuss
Correct answer is : A
Explanation

No of ways of selecting two squares from 64 squares on the chess board is 64C2.

Three cases arise:

 

case 1: when first square is any of the four corner ones

In this case second square can be chosen in 2 ways

no of ways of selecting two squares: 4*2 = 8

 

case2: when first square is any of the 24 squares on the side of the chess board other than the corner ones

the second square can be chosen in 3 ways

no of ways of selecting two squares: 24*3 = 72

 

case 3: the first square is any of the 36 remaining squares

the second square can be chosen in 4 ways

no of ways of selecting two squares: 36*4 = 144

 

Total no of ways of selecting two adjacent squares = 8 + 72 + 144 = 224

 

So required probability = 224/64C2 = 224/4032 = 1/18

 
9

There are ten pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is at least one pair is

A.

195/323

B.

99/323

C.

198/323

D.

185/323

See Answer & Explanation Lets Discuss
Correct answer is : B
Explanation

We will solve this question in a moment. First let us understand one concept which is bit confusing.

Suppose there are 4 objects and I ask you to pick up 2 from them. In how many ways can you do that?

The answer is simple: 4C2 = 6

Can I do like this: 

Out of 4 given objects first I will pick one object(4C1). Then again I will pick one more object in the next draw from the remaining 3 objects(3C1).

This way I have selected two objects from fours objects and number of ways to do this is: 4C1 3C1 = 12

 

Do you see in both the way I am selecting two objects from four objects but in first way of selection the result is 6 while in the other way it is 12. Why so ??

This is because in the second way when we are selecting two objects one by one, their order of selection is being counted in the result. Didn't understand ?

 

OK, let's understand this way: Suppose we mark all 4 objects from 1 to 4.

One way is: In the first draw we select object marked 1 and in the second draw we pick object marked 2.

The other way is:  the first selected object is marked 2 and second object is marked 1.

We are selecting the same two objects but the number of way is 2 because their order of selection (1,2) and (2,1) made it happen so.

But our ultimate goal was to pick up two object and the order does not matter to us.

 

Similarly if you select 3 objects from these 4 objects you can have following orders for objects marked 1,2,3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) = 6 ways= 3! ways. While it should be just 1 way and not 6 ways.

 

So we can conclude that we will have to divide the result by the factorial of the number of objects we are selecting.

Like in the first example when we are selecting 2 objects we will have to divide the result by 2! and in the second example when we are selecting 3 objects we will have to divide the result by 3!.

 

Now coming back to our given problem:

At least on pair of socks must be selected means either one pair or two pair.

So required probability will be: 1 - (no pair of socks are selected)

 

The way we can select no pair of socks is:

Select one sock from 20 sock in the first pick. In the second draw exclude the pair of the first selected sock and pick from the remaining 18 socks and so on.

= 20C1 * 18C1 * 16C1 * 14C1/4! (4! because of the above explained concept)

So required probability: 1 - ((20C1 * 18C1 * 16C1 * 14C1/4!)/20C4) = 99/323

 
10

Ram and Shyam have a cube each. Ram paints four faces of cube with red color and rest with blue color. Ram asks Shyam to paint his cube as well with some sides red and some side blue. They now start rolling the cubes simultaneously. After doing this for very long Ram observes x that probability of both cubes coming up with same color is 1/3. How many faces of his cube did Shyam paint red ?

A.

0

B.

2

C.

3

D.

4

See Answer & Explanation Lets Discuss
Correct answer is : A
Explanation

Let Shyam paints n faces of the cube red. So blue faces of Shyam's cube = (6-n)

Probability that both show up same color is: (probability both show red) + (probability both show blue)

= 4/6 * n/6 + 2/6 * (6-n)/6

= (2n + 6)/18

 

But as given, (2n + 6)/18 = 1/3 => n = 0

 

So Shyam hasn't painted any of the face with red color.

So answer is (a)

 
 

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