Online Mock Tests 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 1
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Subject :- Quantitative Aptitude
Chapter :- Probability-Test 1
Questions :- 25
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A bag contains 8 apple and 6 orange. Four fruits are drawn out one by one and not
replaced. What is the probability that they are alternatively of different fruits?
Correct
Fruits can be drawn in two format
AOAO and OAOA
Apple drawn 1st P=8/14*6/13*7/12*5/11
Orange drawn 1st P=6/14*8/13*5/12*7/11
Adding both we get
2[8*7*6*5/14*13*12*11]=2*(10/143)=20/143.
Incorrect
Fruits can be drawn in two format
AOAO and OAOA
Apple drawn 1st P=8/14*6/13*7/12*5/11
Orange drawn 1st P=6/14*8/13*5/12*7/11
Adding both we get
2[8*7*6*5/14*13*12*11]=2*(10/143)=20/143.
Question 2 of 25
2. Question
1 points
In an interview the probability of Praveen to got selected is 0.4. The probability of
Geetha to got selected is 0.5. The probability of Sam to got selected is 0.6. The probability of Suresh to got selected is 0.8. What is the probability that at least 2 of
them got selected on that day?
Correct
Required probability=1 – no one got
selected – 1 got selected
No one got selected = (1-0.4) x (1-0.5) x
(1-0.6) x (1-0.8) = 0.024
1 got selected= 0.4 x ((1-0.5) x (1-0.6) x (1-
0.8) ) + 0.5 x ((1-0.4) x (1-0.6) x (1-0.8))
+0.6 x ((1-0.4) x (1-0.5) x (1-0.8)) + 0.8 x
((1-0.4) x (1-0.5) x (1-0.6))
= 0.016 + 0.024 + 0.036 + 0.096 = 0.17
So, Required probability = 1 – 0.024 – 0.17= 0.806
Incorrect
Required probability=1 – no one got
selected – 1 got selected
No one got selected = (1-0.4) x (1-0.5) x
(1-0.6) x (1-0.8) = 0.024
1 got selected= 0.4 x ((1-0.5) x (1-0.6) x (1-
0.8) ) + 0.5 x ((1-0.4) x (1-0.6) x (1-0.8))
+0.6 x ((1-0.4) x (1-0.5) x (1-0.8)) + 0.8 x
((1-0.4) x (1-0.5) x (1-0.6))
= 0.016 + 0.024 + 0.036 + 0.096 = 0.17
So, Required probability = 1 – 0.024 – 0.17= 0.806
Question 3 of 25
3. Question
1 points
A basket contains 10 red ball and 15 white ball. out of which 3 red and 4 white balls
are damaged. If two balls selected at random, what is the probability that either
both are white balls or both are not damaged?
A box contains tickets numbered from 1 to 16. 3 tickets are to be chosen to give 3 prizes. What is the probability that at least 2 tickets contain a number which is multiple of 4?
Correct
From 1 to 16, there are 4 numbers which
are multiple of 4
1st 2 are multiple of 4, and one any other
number from (16-4) = 12 tickets
4c2*12c1/16c3 = 72/560
2nd all are multiples of 4.
4c3/16c3=4/560
Add both 72/560+4/560.=76/560=19/140
Incorrect
From 1 to 16, there are 4 numbers which
are multiple of 4
1st 2 are multiple of 4, and one any other
number from (16-4) = 12 tickets
4c2*12c1/16c3 = 72/560
2nd all are multiples of 4.
4c3/16c3=4/560
Add both 72/560+4/560.=76/560=19/140
Question 5 of 25
5. Question
1 points
Chance that Sheela tells truth is 35% and for Ramesh is 75%. In what percent they
likely to contradict each other in the same question?
Correct
P(A) = 35/100=7/20 and P(B) =
75/100=3/4.
Now they are contradicting means one lies
and other speaks truth. So,
Probability = 7/20*1/4 + 13/20 * 3/4
=7/80+39/80=46/80=23/40
Incorrect
P(A) = 35/100=7/20 and P(B) =
75/100=3/4.
Now they are contradicting means one lies
and other speaks truth. So,
Probability = 7/20*1/4 + 13/20 * 3/4
=7/80+39/80=46/80=23/40
Question 6 of 25
6. Question
1 points
Two dice are thrown simultaneously. What is the probability of getting the sum of the
numbers as even?
Correct
Throw two dice n(s)=36
E is nos sum is even.
Hence E={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) ,(2,6 ), …………( 6,2 ), (6,4 ), ( 6,6 )}
n(E)= 18
Thus required probability= 18/36= 1/2
Incorrect
Throw two dice n(s)=36
E is nos sum is even.
Hence E={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) ,(2,6 ), …………( 6,2 ), (6,4 ), ( 6,6 )}
n(E)= 18
Thus required probability= 18/36= 1/2
Question 7 of 25
7. Question
1 points
A basket contains 8 Red and 6 Pink toys. There is another basket which contains 7
Red and 8 Pink toys. One toy is to drawn from either of the two baskets. What is the
probability of drawing a Pink toys?
Correct
Probability of one basket =1/2
1st Basket Pink toy probability =1/2*
(6c1/14c1)
2nd Basket Pink toy probability = 1/2*
(8c1/15c1)
Adding both (1/2*6/14) + (1/2*4/15)
3/14+4/15=101/210
Incorrect
Probability of one basket =1/2
1st Basket Pink toy probability =1/2*
(6c1/14c1)
2nd Basket Pink toy probability = 1/2*
(8c1/15c1)
Adding both (1/2*6/14) + (1/2*4/15)
3/14+4/15=101/210
Question 8 of 25
8. Question
1 points
Four persons are chosen at random from a group of 3 men, 5 women and 4 children.
What is the probability of exactly two of them being men?
Correct
Total People = 3 + 5 + 4 = 12
n(s)=12c4
Probability of exactly two men and two
from others
N(e) = 3c2*9c2
⇒ P= (3c2*9c2)/12c4=>12/55
Incorrect
Total People = 3 + 5 + 4 = 12
n(s)=12c4
Probability of exactly two men and two
from others
N(e) = 3c2*9c2
⇒ P= (3c2*9c2)/12c4=>12/55
Question 9 of 25
9. Question
1 points
A box contains 3 ballons of 1 shape, 4 ballons of 1 shape and 5 ballons of 1 shape. Three ballons of them are drawn at random, what is the probability that all the three are of different shape?
12 persons are seated around a round table.What is the probability that two particular persons sit together?
Correct
In a circle of n different persons, the total
number of arrangements possible = (n – 1)!
n(S) = (12 – 1) = 11 !
Taking two persons as a unit, total persons = 11
Therefore no. of ways for these 11 persons
to around the circular table = (11 – 1)! = 10!
In any unit, 2 particular person can sit in 2! ways.
Hence total number of ways that any three
person can sit,
=n(E) = 10! * 2!
Therefore P (E) = probability of three
persons sitting together = n(E) / n(S)
= (10! * 2!)/11! = 2/11.
Incorrect
In a circle of n different persons, the total
number of arrangements possible = (n – 1)!
n(S) = (12 – 1) = 11 !
Taking two persons as a unit, total persons = 11
Therefore no. of ways for these 11 persons
to around the circular table = (11 – 1)! = 10!
In any unit, 2 particular person can sit in 2! ways.
Hence total number of ways that any three
person can sit,
=n(E) = 10! * 2!
Therefore P (E) = probability of three
persons sitting together = n(E) / n(S)
= (10! * 2!)/11! = 2/11.
Question 11 of 25
11. Question
1 points
A bag contains 6 red, 2 blue and 4 green balls. 3 balls are chosen at random. What is
the probability that at least 2 balls chosen will be red?
Correct
There will be 2 cases
Case 1: 2 red, 1 blue orgreen
Prob. = 6C2 × 6C1 / 12C3 = 9/22
Case 2: all 3 red
Prob. = 6C3 / 12C3 = 2/22
Add the cases, required prob. = 9/22 + 2/22
= 11/22 = 1/2
Incorrect
There will be 2 cases
Case 1: 2 red, 1 blue orgreen
Prob. = 6C2 × 6C1 / 12C3 = 9/22
Case 2: all 3 red
Prob. = 6C3 / 12C3 = 2/22
Add the cases, required prob. = 9/22 + 2/22
= 11/22 = 1/2
Question 12 of 25
12. Question
1 points
Tickets numbered 1 to 250 are in a bag. What is the probability that the ticket
drawn has a number which is a multiple of 4 or 7?
Correct
Multiples of 4 up to 120 = 250/4 = 62
Multiples of 7 up to 120 = 250/7 = 35 (take
only whole number before the decimal part)
Multiple of 28 (4×7) up to 250 = 250/28 =8
So total such numbers are = 62 + 35 – 8 =89
So required probability = 89/250
Incorrect
Multiples of 4 up to 120 = 250/4 = 62
Multiples of 7 up to 120 = 250/7 = 35 (take
only whole number before the decimal part)
Multiple of 28 (4×7) up to 250 = 250/28 =8
So total such numbers are = 62 + 35 – 8 =89
So required probability = 89/250
Question 13 of 25
13. Question
1 points
From a deck of 52 cards, 3 cards are chosen at random. What is the probability that all
are face cards?
Correct
There are 3*4 = 12 face cards in 52 cards
So required probability = 12C3 / 52C3 = 11/1105
Incorrect
There are 3*4 = 12 face cards in 52 cards
So required probability = 12C3 / 52C3 = 11/1105
Question 14 of 25
14. Question
1 points
One 5 letter word is to be formed taking all letters – S, A, P, T and E. What is the
probability that this the word formed will contain all vowels together?
Correct
Total words that can be formed is 5! = 120
Now vowels together:
Take: S, P, T and AE
So their arrangement is 4! * 2! = 48
So required probability = 48/120 = 2/5
Incorrect
Total words that can be formed is 5! = 120
Now vowels together:
Take: S, P, T and AE
So their arrangement is 4! * 2! = 48
So required probability = 48/120 = 2/5
Question 15 of 25
15. Question
1 points
Directions (15-17): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains: 24 green balls. Red balls are 4 more than blue balls. Probability of selecting 1 red ball is 4/13 Bag 2 contains: Total balls are 8 more than 7/13 of balls in bag 1. Probability of selecting 1 red ball is 1/3. The ratio of green balls to blue balls is 1 : 2 Bag 3 contains: Red balls equal total number of green and blue balls in bag 2. Green balls equal total number of green and red balls in bag 2. Probability of selecting 1 blue ball is 3/14.
1 ball each is chosen from bag 1 and bag 2, What is the probability that 1 is red and other blue?
Correct
Let red = x, so blue = x-4
So
x/(24+x+(x-4)) = 4/13
Solve, x = 16
So bag 1: red = 16, green = 24, blue = 12
NEXT:
bag 2: total = 8 + 7/13 * 52 = 36
green and blue = y and 2y. Let red balls = z
So z + y + 2y = 36…………………(1)
Now Prob. of red = 1/3
So z/36 = 1/3
Solve, z = 12
From (1), y = 8
So bag 2: red = 12, green = 8, blue = 16
NEXT:
bag 3: red = 8+16 = 24, green = 12+8 = 20
Blue prob. = 3/14
So a/(24+20+a) = 3/14
Solve, a = 12
So bag 3: red = 24, green = 20, blue = 12
Now probability that 1 is red and other blue::
16/52 * 16/36 + 12/52 * 12/36 = 25/117
Incorrect
Let red = x, so blue = x-4
So
x/(24+x+(x-4)) = 4/13
Solve, x = 16
So bag 1: red = 16, green = 24, blue = 12
NEXT:
bag 2: total = 8 + 7/13 * 52 = 36
green and blue = y and 2y. Let red balls = z
So z + y + 2y = 36…………………(1)
Now Prob. of red = 1/3
So z/36 = 1/3
Solve, z = 12
From (1), y = 8
So bag 2: red = 12, green = 8, blue = 16
NEXT:
bag 3: red = 8+16 = 24, green = 12+8 = 20
Blue prob. = 3/14
So a/(24+20+a) = 3/14
Solve, a = 12
So bag 3: red = 24, green = 20, blue = 12
Now probability that 1 is red and other blue::
16/52 * 16/36 + 12/52 * 12/36 = 25/117
Question 16 of 25
16. Question
1 points
Some green balls are transferred from bag 1 to bag 3. Now probability of choosing a
blue ball from bag 3 becomes 3/16. Find the number of remaining balls in bag 1.
Correct
blue balls in bag 3 are 12
Let x green balls are transferred. So
12/(56+x) = 3/16 [56 was number of bags
in bag 3 before transfer] Solve, x = 8
So remaining number of balls in bag 1 = 52-8 = 44
Incorrect
blue balls in bag 3 are 12
Let x green balls are transferred. So
12/(56+x) = 3/16 [56 was number of bags
in bag 3 before transfer] Solve, x = 8
So remaining number of balls in bag 1 = 52-8 = 44
Question 17 of 25
17. Question
1 points
Green balls in ratio 4 : 1 from bags 1 and 3 respectively are transferred to bag 4. Also 4
and 8 red balls from bags 1 and 3 respectively . Now probability of choosing green ball from bag 4 is 5/11. Find the number of green balls in bag 4?
Correct
4x and x = 5x green balls
4+8 = 12 red balls
So 5x/(5x+12) = 5/11
Solve, x = 2
5*2 = 10 green balls
Incorrect
4x and x = 5x green balls
4+8 = 12 red balls
So 5x/(5x+12) = 5/11
Solve, x = 2
5*2 = 10 green balls
Question 18 of 25
18. Question
1 points
Directions (18-19): There are 3 people – A, B and C. Probability that A speaks truth is 3/10, probability that B speaks truth is 3/7 and probability that C speaks truth is 5/6. For a particular question asked, at most 2 people speak truth. All people answer to a particular question asked.
What is the probability that B will speak truth for a particular question asked?
Correct
In any case B speaks truth. Now at most 2
people speak truth for 1 question
So case 1: B and A speaks truth
Probability = 3/7 * 3/10 * (1-5/6) = 3/140
Case 2: B and C speaks truth
Probability = 3/7 * ( 1- 3/10) * 5/6 = 5/20
Case 3: Only B speaks truth
Probability = 3/7 * ( 1- 3/10) * (1-5/6) =
1/20
Add the three cases = 6/20 + 3/140 =
45/140 = 9/28
Incorrect
In any case B speaks truth. Now at most 2
people speak truth for 1 question
So case 1: B and A speaks truth
Probability = 3/7 * 3/10 * (1-5/6) = 3/140
Case 2: B and C speaks truth
Probability = 3/7 * ( 1- 3/10) * 5/6 = 5/20
Case 3: Only B speaks truth
Probability = 3/7 * ( 1- 3/10) * (1-5/6) =
1/20
Add the three cases = 6/20 + 3/140 =
45/140 = 9/28
Question 19 of 25
19. Question
1 points
A speaks truth only when B does not speak truth, then what is the probability that C
does not speak truth on a question?
Correct
Case 1: B does not speak truth, A speaks
truth
So A speaks truth here
Probability that C does not speak truth =
3/10 * (1 – 3/7) * ( 1- 5/6) = 1/35
Case 2: B speaks truth
So A does not speak truth here
Probability that C does not speak truth = (
1- 3/10) * 3/7 * ( 1- 5/6) = 1/20
So total = 1/35 + 1/20 = 11/140
Incorrect
Case 1: B does not speak truth, A speaks
truth
So A speaks truth here
Probability that C does not speak truth =
3/10 * (1 – 3/7) * ( 1- 5/6) = 1/35
Case 2: B speaks truth
So A does not speak truth here
Probability that C does not speak truth = (
1- 3/10) * 3/7 * ( 1- 5/6) = 1/20
So total = 1/35 + 1/20 = 11/140
Question 20 of 25
20. Question
1 points
There are 100 tickets in a box numbered 1 to 100. 3 tickets are drawn at one by one.
Find the probability that the sum of number on the tickets is odd.
Correct
There will be 4 cases
Case 1: even, even, odd
Prob. = 1/2 × 1/2 × 1/2
Case 2: even, odd, even
Prob. = 1/2 × 1/2 × 1/2
Case 3: odd, even, even
Prob. = 1/2 × 1/2 × 1/2
Case 4: odd, odd, odd
Prob. = 1/2 × 1/2 × 1/2
Add all the cases, required prob. = 1/2
Incorrect
There will be 4 cases
Case 1: even, even, odd
Prob. = 1/2 × 1/2 × 1/2
Case 2: even, odd, even
Prob. = 1/2 × 1/2 × 1/2
Case 3: odd, even, even
Prob. = 1/2 × 1/2 × 1/2
Case 4: odd, odd, odd
Prob. = 1/2 × 1/2 × 1/2
Add all the cases, required prob. = 1/2
Question 21 of 25
21. Question
1 points
There are 4 green and 5 red balls in first bag. And 3 green and 5 red balls in second
bag. One ball is drawn from each bag. What is the probability that one ball will be
green and other red?
Correct
Case 1:first green, second red
Prob. = 4/9 × 5/8 = 20/72
Case 2:first red, second green
Prob. = 5/9 × 3/8 = 15/72
Add the two cases
Incorrect
Case 1:first green, second red
Prob. = 4/9 × 5/8 = 20/72
Case 2:first red, second green
Prob. = 5/9 × 3/8 = 15/72
Add the two cases
Question 22 of 25
22. Question
1 points
A bag contains 2 red, 4 blue, 2 white and 4 black balls. 4 balls are drawn at random,
find the probability that at least one ball is black.
Correct
Prob. (At least 1 black) = 1 – Prob. (None
black)
So Prob. (At least 1 black) = 1 – (8C4/12C4)
= 1 – 14/99
Incorrect
Prob. (At least 1 black) = 1 – Prob. (None
black)
So Prob. (At least 1 black) = 1 – (8C4/12C4)
= 1 – 14/99
Question 23 of 25
23. Question
1 points
Four persons are chosen at random from a group of 3 men, 3 women and 4 children.
What is the probability that exactly 2 of them will be men?
Correct
2 men means other 2 woman and children
So prob. = 3C2 × 7C2 /10C4 = 3/10
Incorrect
2 men means other 2 woman and children
So prob. = 3C2 × 7C2 /10C4 = 3/10
Question 24 of 25
24. Question
1 points
Tickets numbered 1 to 120 are in a bag. What is the probability that the ticket
drawn has a number which is a multiple of 3 or 5?
Correct
Multiples of 3 up to 120 = 120/3 = 40
Multiples of 5 up to 120 = 120/5 = 24 (take
only whole number before the decimal part)
Multiple of 15 (3×5) up to 120 = 120/15 = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = 56/120 = 7/15
Incorrect
Multiples of 3 up to 120 = 120/3 = 40
Multiples of 5 up to 120 = 120/5 = 24 (take
only whole number before the decimal part)
Multiple of 15 (3×5) up to 120 = 120/15 = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = 56/120 = 7/15
Question 25 of 25
25. Question
1 points
There are 2 people who are going to take part in race. The probability that the first
one will win is 2/7 and that of other winning is 3/5. What is the probability that
one of them will win?
Correct
Prob. of 1st winning = 2/7, so not winning
= 1 – 2/7 = 5/7
Prob. of 2nd winning = 3/5, so not winning
= 1 – 3/5 = 2/5
So required prob. = 2/7 * 2/5 + 3/5 * 5/7 = 19/35
Incorrect
Prob. of 1st winning = 2/7, so not winning
= 1 – 2/7 = 5/7
Prob. of 2nd winning = 3/5, so not winning
= 1 – 3/5 = 2/5
So required prob. = 2/7 * 2/5 + 3/5 * 5/7 = 19/35