Online Quiz in Probability For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLA
Probability-Test 3
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Subject :- Quantitative Aptitude
Chapter :- Probability-Test 3
Questions :- 25
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There are three boxes each containing 3 Pink and 5 Yellow balls and also there are 2
boxes each containing 4 Pink and 2 Yellow balls. A Yellow ball is selected at random. Find the probability that Yellow ball is from a box of the first group?
A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas and 5
Bananas are defective. If two fruits selected at random, what is the probability that either
both are Bananas or both are non-defective?
A committee of five persons is to be chosen from a group of 10 people. The probability
that a certain married couple will either serve together or not at all is?
Correct
Five persons is to be chosen from a group of 10
people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126
Incorrect
Five persons is to be chosen from a group of 10
people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126
Question 4 of 25
4. Question
1 points
Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job. The probabilty that atleast one of selected persons will be a Woman is?
Correct
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 –14/91 = 77/91
Incorrect
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 –14/91 = 77/91
Question 5 of 25
5. Question
1 points
Three Bananas and three oranges are kept in a box. If two fruits are chosen at random,
Find the probability that one is Banana and another one is orange?
Correct
Total probability = 6C2 = 15
Probability that one is Banana and another one is
orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5
Incorrect
Total probability = 6C2 = 15
Probability that one is Banana and another one is
orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5
Question 6 of 25
6. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls picked up
random, What is the probability that all three are White?
Correct
Total Balls = 15
Probability = 6c3 / 15c3 = 4/91
Incorrect
Total Balls = 15
Probability = 6c3 / 15c3 = 4/91
Question 7 of 25
7. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls are picked at
random, what is the probability that two are Black and one is Green?
Correct
Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455
Incorrect
Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455
Question 8 of 25
8. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at
random, what is the probability that atleast one is Black?
Correct
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Incorrect
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Question 9 of 25
9. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls.If two balls are picked at
random, what is the probability that either both are Pink or both are Green?
Correct
Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105
Incorrect
Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105
Question 10 of 25
10. 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 11 of 25
11. 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 12 of 25
12. 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 13 of 25
13. 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 14 of 25
14. 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 15 of 25
15. 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 16 of 25
16. 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 17 of 25
17. 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 18 of 25
18. 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 19 of 25
19. 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 20 of 25
20. 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 21 of 25
21. 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 22 of 25
22. 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 23 of 25
23. 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 24 of 25
24. 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 choices and 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 choices and so on)
So, probability = 105/256
Question 25 of 25
25. 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)
=3/25+8/25=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)
