Quant Quiz-Permutation & Combination, Probability
1.How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
A.5
B.10
C.15
D.20
E.None of these
2.Two dice are tossed. The probability that the total score is a prime number is:
A.1/6
B.5/12
C.1/2
D.7/9
E.None of these
3.There are 6 boxes numbered 1, 2,....6. Each box is to be filled up either with a red or a green ball in such a way that at least-
1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
A. 5
B. 21
C. 33
D. 60
E. 6
4.In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?
A.7!
B.7C7
C.77
D.49
E.None of these
5.A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?
A.1
B.1/256
C.81/256
D.175/256
E.144/256
6.Rohit has 9 pairs of dark Blue socks and 9 pairs of Black socks. He keeps them all in a same bag.
If he picks out three socks at random what is the probability he will get a matching pair?
A.(2*9C2 * 9C1) / 18C3
B.(9C2 * 9C1) / 18C3
C.1
D.Cant determine
E.None of these
7.There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the-
albums should contain at least 1 Rock song and 1 Carnatic song?
A.15624
B. 16384
C.6144
D.240
E.None of these
8.From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee.
In how many ways can it be done?
A.564
B.645
C.735
D.756
E.None of these
9.A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box,
if at least one black ball is to be included in the draw?
A.32
B.48
C.64
D.96
E.None of these
10. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A.63
B.90
C.126
D.45
E.135
11.A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
A.10/21
B.11/21
C.2/7
D.5/7
E.None of these
12.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
A.1/3
B3./4
C.7/19
D8/21
E.9/21
13.Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
A.1/2
B.3/4
C.3/8
D.5/16
E.None of these
14.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
A.1/10
B.2/5
C.2/7
D.5/7
E.None of these
15.From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
A.1/15
B.25/57
C.35/256
D.1/221
E.None of these
Answers:-
1.D
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
2.B
3.B
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
4.A
There are seven positions to be filled.
The first position can be filled using any of the 7 letters contained in PROBLEM.
The second position can be filled by the remaining 6 letters as the letters should not repeat.
The third position can be filled by the remaining 5 letters only and so on.
758
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways.
5.D
The probability that he will not hit the target in one shot = 1 - 1/4 = 3/4
Therefore, the probability that he will not hit the target in all the four shots =81/256
Hence, the probability that he will hit the target at least in one of the four shots = 1 - 81/256
= 175/256 .
6.C
If he picks any of the three socks invariably any two of them should match. Hence the probability is 1.
7.A
8.D
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=756
9.C
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
=64
10.A
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1)
=63
11.A
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S)= Number of ways of drawing 2 balls out of 7
= 7C2 `
=(7 x 6)/(2 x 1)
= 21.
Let E = Event of drawing 2 balls, none of which is blue.
n(E)= Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
=(5 x 4)/(2 x 1)
= 10.
P(E) =n(E)/n(S=10/.21
12.A
Total number of balls = (8 + 7 + 6) = 21.
Let E= event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7.
P(E) =n(E)/n(S)=7/21=1/3.
13.B
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) =n(E)/n(S)=27/36=3/4
14.C
P (getting a prize) = 10/(10 + 25)=10/35 =2/7 .
15.D
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
2.B
3.B
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
4.A
There are seven positions to be filled.
The first position can be filled using any of the 7 letters contained in PROBLEM.
The second position can be filled by the remaining 6 letters as the letters should not repeat.
The third position can be filled by the remaining 5 letters only and so on.
758
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways.
5.D
The probability that he will not hit the target in one shot = 1 - 1/4 = 3/4
Therefore, the probability that he will not hit the target in all the four shots =81/256
Hence, the probability that he will hit the target at least in one of the four shots = 1 - 81/256
= 175/256 .
6.C
If he picks any of the three socks invariably any two of them should match. Hence the probability is 1.
7.A
8.D
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=756
9.C
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
=64
10.A
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1)
=63
11.A
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S)= Number of ways of drawing 2 balls out of 7
= 7C2 `
=(7 x 6)/(2 x 1)
= 21.
Let E = Event of drawing 2 balls, none of which is blue.
n(E)= Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
=(5 x 4)/(2 x 1)
= 10.
P(E) =n(E)/n(S=10/.21
12.A
Total number of balls = (8 + 7 + 6) = 21.
Let E= event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7.
P(E) =n(E)/n(S)=7/21=1/3.
13.B
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) =n(E)/n(S)=27/36=3/4
14.C
P (getting a prize) = 10/(10 + 25)=10/35 =2/7 .
15.D
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