Online Mock Tests in Permutation and Combination For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Permutation and Combination-Test 2
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Subject :- Quantitative Aptitude
Chapter :- Permutation and Combination-Test 2
Questions :- 25
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There are 6 men and 7 women. In how many ways a committee of 4 members can be made such that a particular man is always to be excluded?
Correct
There are total 13 people, a particular man is to be excluded, so now 12 people are left to chosen from and
4 members to be chosen. So ways are 12C4.
Incorrect
There are total 13 people, a particular man is to be excluded, so now 12 people are left to chosen from and
4 members to be chosen. So ways are 12C4.
Question 2 of 25
2. Question
1 points
How many 4 digit words can be made from the digits 7, 8, 5, 0, and 4 without repetition?
Correct
0 cannot be on first place for it to be a 4 digit number,
So for 1st digit 4choices, for second also 4 (because 0 can be placed here), then 3 for third place, 2 for
fourth place
Total numbers = 4*4*3*2=96
Incorrect
0 cannot be on first place for it to be a 4 digit number,
So for 1st digit 4choices, for second also 4 (because 0 can be placed here), then 3 for third place, 2 for
fourth place
Total numbers = 4*4*3*2=96
Question 3 of 25
3. Question
1 points
In how many ways 8 students can be given 3 prizes such that no student receives more than 1 prize?
Correct
For 1st prize there are 8 choices, for 2nd prize, 7 choices, and for 3rd prize – 6 choices left
So total ways = 8*7*6=336
Incorrect
For 1st prize there are 8 choices, for 2nd prize, 7 choices, and for 3rd prize – 6 choices left
So total ways = 8*7*6=336
Question 4 of 25
4. Question
1 points
A box contains 27 marbles some are blue and others are green. If a marble is drawn at random from the box, the probability that it is blue is 1/3. Then how many number of green marbles in the box?
Correct
Blue marble – x
xc1/27c1 = 1/3
x/27=1/3 —> x=27/3=9
No of green marbles = Total – Blue marble =27-9=18
Incorrect
Blue marble – x
xc1/27c1 = 1/3
x/27=1/3 —> x=27/3=9
No of green marbles = Total – Blue marble =27-9=18
Question 5 of 25
5. Question
1 points
In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random.What is the probability that at least one ball is of red colour?
Correct
Total Balls =10
Other than red ball = 6c2
6c2/10c2=1/3 —> 1-1/3 = 2/3
Incorrect
Total Balls =10
Other than red ball = 6c2
6c2/10c2=1/3 —> 1-1/3 = 2/3
Question 6 of 25
6. Question
1 points
Sahil has two bags (A & B) that contain green and blue balls.In the Bag ‘A’ there are 6 green and 8 blue balls and in the Bag ‘B’ there are 6 green and 6 blue balls. One ball is drawn out from any of these two bags. What is the probability that the ball drawn is blue?
Correct
Total balls in A bag = 14, Total balls in A bag = 12
A bag = 1/2(8c1/14c1) = 2/7
B bag = 1/2(6c1/12c1) = 1/4 —> total Probability = 2/7 + 1/4 =15/28
Incorrect
Total balls in A bag = 14, Total balls in A bag = 12
A bag = 1/2(8c1/14c1) = 2/7
B bag = 1/2(6c1/12c1) = 1/4 —> total Probability = 2/7 + 1/4 =15/28
Question 7 of 25
7. Question
1 points
In an examination, there are three sections namely Reasoning, Maths and English. Reasoning part contains 4 questions. There are 5 questions in maths section and 6 questions in English section.If three questions are selected randomly from the list of questions then what is the probability that all of them are from maths?
Correct
Total no of questions= 15
Probability = 5c3/15c3 = 2/91
Incorrect
Total no of questions= 15
Probability = 5c3/15c3 = 2/91
Question 8 of 25
8. Question
1 points
A basket contains 5 red 4 blue 3 green marbles. If three marbles picked up random, What is the probability that either all are green or all are red?
Correct
Total Marbles = 12
Either all are green or all are red = 5c3 + 3c3
probability = 5c3 + 3c3/12c3 = 11/220 = 1/20
Incorrect
Total Marbles = 12
Either all are green or all are red = 5c3 + 3c3
probability = 5c3 + 3c3/12c3 = 11/220 = 1/20
Question 9 of 25
9. Question
1 points
A basket contains 5 red 4 blue 3 green marbles. If three marbles picked up random, What is the probability that at least one is blue?
Correct
Total Marbles = 12
other than blue 8c3 / 12c3 = 14/55
probability = 1-14/55 = 41/55
Incorrect
Total Marbles = 12
other than blue 8c3 / 12c3 = 14/55
probability = 1-14/55 = 41/55
Question 10 of 25
10. Question
1 points
A basket contains 5 red 4 blue 3 green marbles. If two marbles picked up random, What is the probability that both are red?
Correct
Total Marbles = 12
Probability = 5c2 / 12c2 = 5/33
Incorrect
Total Marbles = 12
Probability = 5c2 / 12c2 = 5/33
Question 11 of 25
11. Question
1 points
A bag contains 5 red caps, 4 blue caps, 3 yellow caps and 2 green caps.If three caps are picked at random, what is the probability that two are red and one is green?
Correct
Total caps = 14
Probability = 5c2 * 2c1/ 14c3 = 5/91
Incorrect
Total caps = 14
Probability = 5c2 * 2c1/ 14c3 = 5/91
Question 12 of 25
12. Question
1 points
A bag contains 5 red caps, 4 blue caps, 3 yellow caps and 2 green caps. If four caps are picked at random, what is the probability that two are red, one is blue and one is green?
Correct
Total caps = 14
Probability = 5c2 * 4c1 * 2c1 / 14c4 = 80/1001
Incorrect
Total caps = 14
Probability = 5c2 * 4c1 * 2c1 / 14c4 = 80/1001
Question 13 of 25
13. Question
1 points
A bag contains 2 red caps, 4 blue caps, 3 yellow caps and 5 green caps. If three caps are picked at random, what is the probability that none is green?
Correct
Total caps = 14
Probability = 5c0 * 9c3/ 14c3 = 3/13
Incorrect
Total caps = 14
Probability = 5c0 * 9c3/ 14c3 = 3/13
Question 14 of 25
14. Question
1 points
A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?
Correct
Balls are picked in two manners – RWRW or WRWR
So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99
Incorrect
Balls are picked in two manners – RWRW or WRWR
So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99
Question 15 of 25
15. Question
1 points
P and Q are sitting in a ring with 11 other persons. If the arrangement of 11 persons is at
random, then the probability that there are exactly 4 persons between them?
Correct
Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12
Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so
favourable outcomes = 2
So, probability = 2/12 = 1/6
Incorrect
Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12
Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so
favourable outcomes = 2
So, probability = 2/12 = 1/6
Question 16 of 25
16. Question
1 points
10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?
Correct
Total outcomes = (10 -1)! = 9!
Favourable outcomes = 6!*4! (4 person seated together and 6 other persons seated randomly, so they will
sit in (7-1)! Ways and those 4 persons can be arranged in 4! ways)
So probability = 1/21
Incorrect
Total outcomes = (10 -1)! = 9!
Favourable outcomes = 6!*4! (4 person seated together and 6 other persons seated randomly, so they will
sit in (7-1)! Ways and those 4 persons can be arranged in 4! ways)
So probability = 1/21
Question 17 of 25
17. Question
1 points
A box contains 4 red, 5 black and 6 green balls. 3 balls are drawn at random. What is the
probability that all the balls are of same colour?
Correct
(4c3 + 5c3 + 6c3)/15c3 = 34/455
Incorrect
(4c3 + 5c3 + 6c3)/15c3 = 34/455
Question 18 of 25
18. Question
1 points
An apartment has 8 floors. An elevator starts with 4 passengers and stops at 8 floors of the apartment. What is the probability that all passengers travels to different floors?
Correct
Total outcomes = 8*8*8*8
Favourable outcomes = 8*7*6*5 (first person having 8 choices, after that second person have 7 choicesand so on)
So, probability = 105/256
Incorrect
Total outcomes = 8*8*8*8
Favourable outcomes = 8*7*6*5 (first person having 8 choices, after that second person have 7 choicesand so on)
So, probability = 105/256
Question 19 of 25
19. Question
1 points
A speak truth in 60% cases and B in 80% cases. In what percent of cases they likely to
contradict each other narrating the same incident?
Correct
P(A) = 3/5 and P(B) = 4/5. Now they are contradicting means one is telling truth and other telling the lie.
So,
Probability = (3/5)*(1/5) + (2/5)*(4/5)=11/25
Incorrect
P(A) = 3/5 and P(B) = 4/5. Now they are contradicting means one is telling truth and other telling the lie.
So,
Probability = (3/5)*(1/5) + (2/5)*(4/5)=11/25
Question 20 of 25
20. Question
1 points
A box contains 30 electric bulbs, out of which 8 are defective. Four bulbs are chosen at random from this box. Find the probability that at least one of them is defective?
Correct
1 – 22c4/30c4 = 1 – 209/783 = 574/783
Incorrect
1 – 22c4/30c4 = 1 – 209/783 = 574/783
Question 21 of 25
21. Question
1 points
Two person A and B appear in an interview. The probability of A’s selection is 1/5 and the
probability of B’s selection is 2/7. What is the probability that only one of them is selected?
Correct
A selects and B rejects + B selects and A rejects = (1/5)*(5/7) + (4/5)*(2/7) = 13/35
Incorrect
A selects and B rejects + B selects and A rejects = (1/5)*(5/7) + (4/5)*(2/7) = 13/35
Question 22 of 25
22. Question
1 points
A 4- digit number is formed by the digits 0, 1, 2, 5 and 8 without repetition. Find the probability that the number is divisible by 5.
Correct
Total possibility = 5*4*3*2
Favourable outcomes = 2*4*3*2 (to be divisible by 5 unit digit can be filled with only 0 or 5, so only two possibilities are there, then the remaining can be filled in 4, 3 and 2 ways respectively)
so probability = 2/5
Incorrect
Total possibility = 5*4*3*2
Favourable outcomes = 2*4*3*2 (to be divisible by 5 unit digit can be filled with only 0 or 5, so only two possibilities are there, then the remaining can be filled in 4, 3 and 2 ways respectively)
so probability = 2/5
Question 23 of 25
23. Question
1 points
A bag contains 6 red balls and 8 green balls. 2 balls are drawn at random one by one with
replacement. Find the probability that both the balls are green
Correct
(8c1)/(14c1) * (8c1)*(14c1) = 16/49
Incorrect
(8c1)/(14c1) * (8c1)*(14c1) = 16/49
Question 24 of 25
24. Question
1 points
In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes ?
Correct
12^3 = 1728
Incorrect
12^3 = 1728
Question 25 of 25
25. Question
1 points
Total no of ways in which 30 sweets can be distributed among 6 persons ?