Probability and Statistics - Online Test
Q1. If the mean of the first n odd natural numbers be n itself, then n is equal to
Answer : Option C
Explaination / Solution:
here, n can be any natural number
Q2. If P(A) = , P(B) = and P(A ∪ B) = find P(A|B)
Answer : Option B
Explaination / Solution:
Q3.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at least one is green?
Answer : Option B
Explaination / Solution:
Probability if at least one is Green
[1-(9C2/14C2)] = 55/91
Q4. The mean of the first n terms of the A.P. (a + d ) + ( a + 3d ) + ( a + 5d ) +………..is
Answer : Option B
Explaination / Solution:
Q5.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that either both are red or both are green?
Answer : Option D
Explaination / Solution:
Probabilities if both either are Red or either are green
(5C2 + 5C2)/ 14C2 = (10+10)/91 --> 20/91
Q6. Two dice are thrown. The number of sample points in the sample space when 6 does not appear on either dice is
Answer : Option C
Explaination / Solution:
total no. of outcomes=62=36
The pair which shows 6 on either dice are {(6,1),(6,2),(6,3),(6,4),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}=11
So total number of sample points in sample space when 6 on either dice do not appear=36-11=25
Q7. If a, b, c be any three positive numbers, then the least value of
Answer : Option A
Explaination / Solution:
A.M≥H.M⇒a+b+c3≥31a+1b+1c⇒(a+b+c)(1a+1b+1c)≥3×3≥9
Q8.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at most one is blue?
Answer : Option B
Explaination / Solution:
Probabilities if at most one is blue =
[(4C0*10C2+ 4C1*10C1)/14C2] = (1*45 + 4*10)/( 91) = 85/91
Q9. 8 coins are tossed at a time. The probability of getting 6 heads up is
Answer : Option B
Explaination / Solution:
Total ways of getting 6 heads out of 8 toss of coins is 28.
Total number of outcome is 28 = 256
Therefore probability is
Q10. The median of the data 13,14,16,18,20,22 is
Answer : Option C
Explaination / Solution:
