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12/16/2014 eLitmus Probability questions for eLitmus | eLitmus Questions for pH Test Page 1
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Elitmus Question Paper
Probability
eLitmus Logical Reasoning
Questions
Number System
Time and Work
Geometry
Permutation and
Combination
:: Topics :: For more details on eLitmus click Here:
eLitmus Syllabus , Quant sample Questions and Cryptarithmetic Questions
1 What is the probability of getting at least one six in a single throw of three unbiased
dice ?
A. 31/216
B. 91/216
C. 125/216
D. 81/216
See Answer & Explanation Lets Discuss
Correct answer is : B
Explanation
Three cases arise:
Case 1: When only one dice shows up a six
This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three
independent events and add them up to get the total probability
Probability that only 1st dice shows up a six: (probability that first dice shows up a
6) and (probability that second dice shows up other than 6) and (probability that
third dice shows up other than 6)
=(1/6)*(5/6)*(5/6)
=25/216
similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216
And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216
So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) =
75/216
Case 2:When two dice show up a six
Total number of ways of selecting a pair of dice that show up a six from a set of 3
dice are: 3C2=3
Find the probability of getting six on a pair of dice and multiply it by total number of
such possible pairs
Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216
So, total probability = 3*(5/216) = 15/216
Case 3: When all dice show up a six
In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216
So total probability of getting at least one six = (75/216) + (15/216) + (1/216) =
91/216
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What is the probability that a two digit number selected at random is a multiple of 3
and not a multiple of 5 ?
A. 1/15
B. 2/15
C. 4/15
D. 7/15
See Answer & Explanation Lets Discuss
Correct answer is : C
Explanation
Since every third number starting from 10 will be divisible by 3, so total number of
numbers divisible by 3 are 90/3 = 30
Numbers which are divisible by 3 and 5 both are numbers which are multiple of 15.
For the range 10 to 99, 15 is the first number divisible by 15 and 90 is the last
number.
So total number of numbers divisible by 15 are: (90-15)/15 + 1 = 5 + 1 = 6
Number of numbers which are divisible by 3 are 30 and number of numbers which are
divisible by 3 and 5 both are 6. So number of numbers divisible by 3 and not by 5 are:
30-6=24
So total probability = 24/90 = 4/15
3 Out of 40 consecutive integers, two are chosen at random. Find the probability that
their sum is odd?
A. 21/40
B. 19/39
C. 21/39
D. 20/39
See Answer & Explanation Lets Discuss
Correct answer is : D
Explanation
Forty consecutive integers will have 20 even and 20 odd integers. The sum of 2
chosen integers will be odd, only if:
First is even and second is odd
First is odd and second is even
Mathematically the probability will be given by:
P(First is even) * P(Second is odd) + P(First is odd) + P(Second is even)
= (20/40) * (20/39) + (20/40) * (20/39)
= 20/39
4 8 coins are tossed. Whats is the probability that one and only one turns up head ?
A. 1/256
B. 1/32
C. 1/64
D. 1/128
See Answer & Explanation Lets Discuss
Correct answer is : B
Explanation
The coin that turns up head can be any of the eight.
Required probability = 8 * (1/28)
= 1/32
5 A fair coin is tossed repeatedly. If head appears on first 5 tosses what is the
probability that tail appears on 6th toss ?
A. 1/6
B. 1/5
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1 2 3 4 5 6
C. 1/3
D. 1/2
See Answer & Explanation Lets Discuss
Correct answer is : D
Explanation
Like mentioned in the question, the coin is fair. So chance of appearing of head and
tail on each toss is same and each toss is independent from the previous one. So the
chance of appearing tail on 6th toss is still 1/2.
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