Directions: Each of the questions given below is followed by series of options among which you have to choose the best one. You can also review your choices by clicking into answer section.
16. A bag contains 4 blue, 5 white and 6 green balls. Two balls are drawn at random. What is the probability that both the balls are blue?
A. 2/35
B. 1/17
C. 1/15
D. 2/21
Answer
Explanation:
Let E be the event that both the balls are blue.
n(E) = Number of ways in which 2 balls can be drawn from 4 blue balls = 4C2
n(S) = Number of ways in which 2 balls can be drawn from 15 balls (∵ 4 + 5 + 6 = 15) = 15C2
P(E) = n(E)/n(S)=4C2/15C2
=2/35
17. Two dice are thrown simultaneously. Find the probability of getting a prime number on both the dice.
A. 1/4
B. 2/9
C. 1/6
D. 1/5
Answer
Explanation:
n(S) = 36
n(E) = {(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5)}/ n(E) = 9
P(E) = n(E)/n(S)=9/36 = 1/4
18. What is the probability of getting at most two tails in tossing three fair coins?
A. 1/8
B. 7/8
C. 5/8
D. 3/4
Answer Ans: Option B
Explanation:
Probability of getting three tails = 1/8
Probability of getting at most two tails = 1-(1/8) = 7/8
19. When two dice are rolled, what is the probability that the sum is either 7 or 11?
A. 14
B. 25
C. 19
D. 29
Answer
Explanation:
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
Hence, total number of outcomes possible when two dice are rolled = 6 × 6 = 36
To get a sum of 7, the following are the favorable cases. (1, 6), (2, 5), {3, 4}, (4, 3), (5, 2), (6,1)
=Number of ways in which we get a sum of 7 = 6
P (a sum of 7) = Number of ways in which we get a sum of 7
Total number of outcomes possible=6/36
To get a sum of 11, the following are the favorable cases (5, 6), (6, 5)
=Number of ways in which we get a sum of 11 = 2
P(a sum of 11) = Number of ways in which we get a sum of 11
Total number of outcomes possible=2/36
Here, clearly the events are mutually exclusive events.
By Addition Theorem of Probability, we have
P (a sum of 7 or a sum of 11) = P(a sum of 7) + P( a sum of 11)
=(6/36)+(2/36)
=8/36 = 2/9
20. A letter is randomly taken from English alphabets. What is the probability that the letter selected is not a vowel?
A. 525
B. 225
C. 5/26
D. 21/26
Answer
Explanation:
Total number of alphabets, n(S) = 26
Total number of characters which are not vowels, n(E) = 21
P(E) = n(E)/n(S)=21/26
21. Tickets numbered 1 to 100 are placed in a box. Find the probability that a ticket selected at random has a number that is divisible by both 3 and 5.
A. 3/20
B. 3/50
C. 1/15
D. 3/25
Answer
Explanation:
For the number to be divisible by both 3 and 5, it should be a multiple of 15.
E= {15, 30, 45, 60,75, 90}
n(E) = 6
n(S) = 100
P(E) = 6/100 = 3/50
22. Two brothers appeared for an exam. The probability of each of them getting selected is 1/4 and 1/6 respectively. Find the probability that only one of the two is selected.
A. 1/24
B. 1/12
C. 10/24
D. 1/3
Answer
Explanation:
Probability that only of the two are selected = P(A alone selected)+P(B alone selected)
= P(A selected)*P(B not selected)+ P(A not selected)*P(B selected)
= [(1/4)*(5/6)]+[(3/4)*(1/6)] = [5/24]+[3/24] = 8/24 = 1/3
23. Two integers are randomly selected from the below sets, one from A and the other from set B. Find the probability that the sum of the two integers is 11. A={1,2,3,4,5} B={6,7,8,9,10}
A. 1/5
B. 1/10
C. 1/20
D. 1/4
Answer
Explanation:
No. of ways of selecting 2 integers = n(selecting 1 number from A.* n(selecting 1 number from B.
= 5*5 = 25
{1, 10}, {2,9}, (3,8}, (4,7), {5,6} These 5 pairs of integers have their sum as 11.
n(E) = 5/25 = 1/5
