# Elitmus Probability Questions & Answers

**1.(100)power 8-(111112)power 2 divided by 3 power n. find the maximum value of n?**

A)3

B)2

C)1

D)0

**Correct Answer : A**

**Explanation:**

since it is 100^8 converting it in the power of 2.

so 100^8=(100000000)^2 now, 100000000^2-111112^2 .’. a^2-b^2=(a+b)(a-b) =(100000000-111112)(100000000+111112) =99888888*100111112 99888888*100111112

so 9+9+8+8+8+8+8+8=66 66 is divisible by 3 but not 9 so ans should be ->3^1.

**2.In a particular state, the god of rain announces that there will be raining on exactly 20 days in each calender year. if the probability of raining on a particular day of the year 2011, 2012 and 2013 are P, Q, and R respectively then which of the following relation holds true:**

A)2Q>=P+R

B)Q^2 >=P*R

C)2/Q >= 1/P + 1/Q

D)all of these

**Correct Answer : C**

**Explanation:**

Probability of raining on a particular day is 20/365(No. of favorable days/total no of days), for a leap year its probability is 20/266.

Now we have P=20/365, Q=20/366, R=20/365…. now verify options, only option (3) is correct

**3.There are 10 pairs 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

**Correct Answer : B**

**Explanation:**

probability is 1-p(drawing four socks which ar different) p(drawing four socks different) (20/20)*(18/19)*(16/18)*(14/17)=>224/323

so probability is 1-224/323=>99/323

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

A)1/18

B)64/4032

C)63/64

D)1/9

**Correct Answer : A**

**Explanation:**

In 64 squares, there are:

(1) 4 at-corner squares, each has ONLY 2 squares each having a side in common with…

(2) 6*4 = 24 side squares, each has ONLY 3 squares such that each has a side in common with…

(3) 6*6 = 36 inner squares, each has 4 squares such that each has a side in common with…

So we have the calculation: P = (4/64)*(2/63) + (24/64)*(3/63)+ (36/64)*(4/63) P = 1/18

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

A)1

B)1/256

C)81/256

D)175/256

**Correct Answer : D**

**Explanation:**

In four shots he can hit once,twice,thrice,all hit the probability of hitting the target isp(1hit out of 4)+P(2hit out of 4)+p(3hit out of 4)+p(All hit) it is total probability-probability of not hitting yhr target =>1-(3/4*3/4*3/4*3/4) =>175/256

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

A)412/455

B)434/455

C)449/455

D)438/455

**Correct Answer : A**

**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. So answer could be 412/455.

**7.Two pyramids are thrown with faces 1,2,4,6. Find the probability to get a sum of the even number on their faces?**

A)3/4

B)5/8

C)7/8

D)5/4

**Correct Answer : B**

**Explanation:**

1 pyramid can show 4 different kind of arrangements 1,2,4=7 1,2,6=9 1,4,6=11 2,4,6=12

So, out of 4 outcomes 3 are odd and 1 is even.

In question given two pyramids are thrown. So, to make some of their faces as even

Condition 1: some of both individual faces of each pyramid is odd 3/4 * 3/4 =9/16

Condition 2: some of both individual faces of pyramid is even 1/4 + 1/4 =1/16

Total such conditions = 9/16 + 1/16 = 10/16 = 5/8.

**8.The probability of Cristiano Ronaldo scoring a penalty is twice than the probability of missing it. In a World Cup he takes 5 penalties, What is the probability of scoring 3 penalties among the 5?**

A)5*2*((2/3)^3 * (1/3)^3)

B)1-((2/3)^3 * (1/3)^3)

C)1-((2/3)^2)* (1/3)^3)

D)None of these

**Correct Answer : A**

**Explanation:**

P(s)=2/3, P(m)=1/3 The probability of scoring 3 = 5c3*(2/3)^3*(1/3)^2 =5*2*((2/3)^3*(1/3)^2)

**9.P is an even number (I think greater than 2 it was given). e is the probability of a number of even numbers divisible by p, o is the probability of the number of an odd number divisible by p and w is the probability of the number of whole numbers divisible by p. Then which equation satisfies.**

A)w=2 e

B)w=e/2

C)w=o/2

D)None of these

**Correct Answer : B**

**Explanation:**

it was p>99 since p is an even number..

so it can,t be divisible by odds e is the probability of p to be divisible by even numbers..

means there are only even numbers in sample space..

n for w all whole numbers are allowed(odds also) the possible number of values of p are same in both cases but sample space(for w)=2(sample space for e)

so it is clear that w=e/2.

**10.if |x -4 | + |y – 4| =4 then how many integers value the sets (x,y) have?**

A)infinite

B)5

C)16

D)25

**Correct Answer : C**

**Explanation:**

mod of x-4 should be 0 to 4 and also mod of y-4 should be 0 to 4.

So first think all the possible value of x so that x-4=0 and think all the possible value of so that y-4=4.So to make |x-4|=0 and |y-4|=4 value of x and y is 1.(4,0),2.(4,8).

simillarly to make |y-4|=0 and |x-4|=4 value of (x,y) is 3.(0,4),4.(0,8).to make |x-4|=1 and |y-4|=3,

values are 5.(3,1),6.(3,7),7.(5,1),8.(5,7) and |x-4|=3,|y-4|=1

values are9.(1,3),10.(7,3),11.(1,5),12.(7,5).for |x-2|=2 and |y-2|=2

values are 13.(2,2),14.(2,6),15.(6,6),16.(6,2).

so total 16.