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 2
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Subject :- Quantitative Aptitude
Chapter :- Probability-Test 2
Questions :- 25
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Two cards are drawn at random from a pack of 52 cards. What is the probability
that both the cards drawn are face card (Jack, Queen and King)?
Correct
There are 52 cards, out of which there are
12 face cards.
So probability of 2 face cards = 12C2/52C2 = 11/221
Incorrect
There are 52 cards, out of which there are
12 face cards.
So probability of 2 face cards = 12C2/52C2 = 11/221
Question 2 of 25
2. Question
1 points
A committee of 5 people is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have less number of boys than girls?
Correct
Case 1: 1 boy and 4 girls
Prob. = 5C1 × 4C4/9C5 = 5/146
Case 2: 2 boys and 3 girls
Prob. = 5C2 ×
4C3/9C5 = 40/126
Add the two cases = 45/126 = 5/12
Incorrect
Case 1: 1 boy and 4 girls
Prob. = 5C1 × 4C4/9C5 = 5/146
Case 2: 2 boys and 3 girls
Prob. = 5C2 ×
4C3/9C5 = 40/126
Add the two cases = 45/126 = 5/12
Question 3 of 25
3. Question
1 points
A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at
random. What is the probability that they are not of same color?
Correct
Three cases
Case 1: one red, 1 blue
Prob = 2C1 × 4C1 / 12C2 = 4/33
Case 2: one red, 1 white
Prob = 2C1 × 6C1 / 12C2 = 2/11
Case 3: one white, 1 blue
Prob = 6C1 × 4C1 / 12C2 = 4/11
Add all cases
Incorrect
Three cases
Case 1: one red, 1 blue
Prob = 2C1 × 4C1 / 12C2 = 4/33
Case 2: one red, 1 white
Prob = 2C1 × 6C1 / 12C2 = 2/11
Case 3: one white, 1 blue
Prob = 6C1 × 4C1 / 12C2 = 4/11
Add all cases
Question 4 of 25
4. Question
1 points
A bag contains 5 blue balls, 4 black balls and 3 red balls. Six balls are drawn at
random. What is the probability that there are equal numbers of balls of each color?
Correct
5C2× 4C2× 3C2/ 12C6
Incorrect
5C2× 4C2× 3C2/ 12C6
Question 5 of 25
5. Question
1 points
Directions (5-7): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is
2/5. There are 10 blue balls.
What is the probability of choosing one blue ball?
Correct
Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3
Incorrect
Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3
Question 6 of 25
6. Question
1 points
Directions (5-7): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is
2/5. There are 10 blue balls.
What is the total number of balls in the urn?
Correct
Probability of choosing one blue ball is 1/3
And total blue balls are 10. So with 10/30
we get probability as 1/3
So total balls must be 30
Incorrect
Probability of choosing one blue ball is 1/3
And total blue balls are 10. So with 10/30
we get probability as 1/3
So total balls must be 30
Question 7 of 25
7. Question
1 points
Directions (5-7): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is
2/5. There are 10 blue balls.
If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 2 or 3?
Correct
There are 30 balls in the urn.
Multiples of 2 up to 30 = 30/2 = 15
Multiples of 3 up to 30 = 30/3 = 10 (take
only whole number before the decimal part)
Multiples of 6 (2×3) up to 30 = 30/6 = 5
So total such numbers are = 15 + 10 – 5 = 20
So required probability = 20/30 = 2/3
Incorrect
There are 30 balls in the urn.
Multiples of 2 up to 30 = 30/2 = 15
Multiples of 3 up to 30 = 30/3 = 10 (take
only whole number before the decimal part)
Multiples of 6 (2×3) up to 30 = 30/6 = 5
So total such numbers are = 15 + 10 – 5 = 20
So required probability = 20/30 = 2/3
Question 8 of 25
8. Question
1 points
There are 2 brothers A and B. Probability that A will pass in exam is 3/5 and that B
will pass in exam is 5/8. What will be the probability that only one will pass in the
exam?
Correct
Only one will pass means the other will fail
Probability that A will pass in exam is 3/5.
So Probability that A will fail in exam is 1– 3/5 = 2/5
Probability that B will pass in exam is 5/8.
So Probability that B will fail in exam is 1 – 5/8 = 3/8
So required probability = P(A will
pass)*P(B will fail) + P(B will pass)*P(A will fail)
So probability that only one will pass in the
exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40
Incorrect
Only one will pass means the other will fail
Probability that A will pass in exam is 3/5.
So Probability that A will fail in exam is 1– 3/5 = 2/5
Probability that B will pass in exam is 5/8.
So Probability that B will fail in exam is 1 – 5/8 = 3/8
So required probability = P(A will
pass)*P(B will fail) + P(B will pass)*P(A will fail)
So probability that only one will pass in the
exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40
Question 9 of 25
9. Question
1 points
If three dices are thrown simultaneously, what is the probability of having a same
number on all dices?
Correct
Total events will be 6*6*6 = 216
Favorable events for having same number
is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4},
{5,5,5}, {6,6,6} – so 6 events
Probability of same number on all dices is
6/216 = 1/36
Incorrect
Total events will be 6*6*6 = 216
Favorable events for having same number
is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4},
{5,5,5}, {6,6,6} – so 6 events
Probability of same number on all dices is
6/216 = 1/36
Question 10 of 25
10. Question
1 points
There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?
Correct
Multiples of 3 up to 150 = 150/3 = 50
Multiples of 7 up to 150 = 150/7 = 21 (take
only whole number before the decimal part)
Multiples of 21 (3×7) up to 150 = 150/21 = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = 64/150 = 32/75
Incorrect
Multiples of 3 up to 150 = 150/3 = 50
Multiples of 7 up to 150 = 150/7 = 21 (take
only whole number before the decimal part)
Multiples of 21 (3×7) up to 150 = 150/21 = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = 64/150 = 32/75
Question 11 of 25
11. Question
1 points
There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a
ticket which has a prime number on it?
Correct
Prime numbers up to 55 is 16 numbers
which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 43, 41, 47, 53.
So probability = 16/55
Incorrect
Prime numbers up to 55 is 16 numbers
which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 43, 41, 47, 53.
So probability = 16/55
Question 12 of 25
12. Question
1 points
A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue
balls. What is the probability of choosing two balls such that one is white and the
other is blue?
Correct
Case 1: Ball from first bag is white, from
another is blue
So probability = 4/9 * 7/12 = 28/108
Case 1: Ball from first bag is blue, from
another is white
So probability = 5/9 * 5/12 = 25/108
Add the cases
So required probability = 28/108 + 25/108
= 53/108
Incorrect
Case 1: Ball from first bag is white, from
another is blue
So probability = 4/9 * 7/12 = 28/108
Case 1: Ball from first bag is blue, from
another is white
So probability = 5/9 * 5/12 = 25/108
Add the cases
So required probability = 28/108 + 25/108
= 53/108
Question 13 of 25
13. Question
1 points
The odds against an event are 2 : 3 and the odds in favor of another independent event
are 3 : 4. Find the probability that at least one of the two events will occur.
Correct
Let 2 events A and B
Odds against A are 2 : 3
So probability of occurrence of A = 3/(2+3)
= 3/5. And non-occurrence of A = 2/5
Odds in favor of B are 3 : 4
So probability of occurrence of B = 3/(3+4)=3/7
And non-occurrence of B = 4/7
Probability that at least one occurs
Case 1: A occurs and B does not occur
So probability = 3/5 * 4/7 = 12/35
Case 2: B occurs and A does not occur
So probability = 3/7 * 2/5 = 6/35
Case 3: Both A and B occur
So probability = 3/5 * 3/7 = 9/35
So probability that at least 1 will occur =
12/35 + 6/35 + 9/35 = 27/35
Incorrect
Let 2 events A and B
Odds against A are 2 : 3
So probability of occurrence of A = 3/(2+3)
= 3/5. And non-occurrence of A = 2/5
Odds in favor of B are 3 : 4
So probability of occurrence of B = 3/(3+4)=3/7
And non-occurrence of B = 4/7
Probability that at least one occurs
Case 1: A occurs and B does not occur
So probability = 3/5 * 4/7 = 12/35
Case 2: B occurs and A does not occur
So probability = 3/7 * 2/5 = 6/35
Case 3: Both A and B occur
So probability = 3/5 * 3/7 = 9/35
So probability that at least 1 will occur =
12/35 + 6/35 + 9/35 = 27/35
Question 14 of 25
14. Question
1 points
The odds against an event are 1 : 3 and the odds in favor of another independent event
are 2 : 5. Find the probability that one of the event will occur.
Correct
Let 2 events A and B
Odds against A are 1 : 3
So probability of occurrence of A = 3/(1+3)
= 3/4. And non-occurrence of A = 1/4
Odds in favor of B are 2 : 5
So probability of occurrence of B = 2/(2+5)
= 2/7. And non-occurrence of B = 5/7
Case 1: A occurs and B does not occur
So probability = 3/4 * 5/7 = 15/28
Case 2: B occurs and A does not occur
So probability = 2/7 * 1/4 = 2/28
So probability that one will occur = 15/28 +
2/28 = 17/28
Incorrect
Let 2 events A and B
Odds against A are 1 : 3
So probability of occurrence of A = 3/(1+3)
= 3/4. And non-occurrence of A = 1/4
Odds in favor of B are 2 : 5
So probability of occurrence of B = 2/(2+5)
= 2/7. And non-occurrence of B = 5/7
Case 1: A occurs and B does not occur
So probability = 3/4 * 5/7 = 15/28
Case 2: B occurs and A does not occur
So probability = 2/7 * 1/4 = 2/28
So probability that one will occur = 15/28 +
2/28 = 17/28
Question 15 of 25
15. Question
1 points
From a pack of 52 cards, 1 card is chosen at random. What is the probability of the card
being diamond or queen?
Correct
In 52 cards, there are 13 diamond cards and 4
queens.
1 card is chosen at random
For 1 diamond card, probability = 13/52
For 1 queen, probability = 4/52
For cards which are both diamond and queen,
probability = 1/52
So required probability = 13/52 + 4/52 – 1/52 =
16/52 = 4/13
Incorrect
In 52 cards, there are 13 diamond cards and 4
queens.
1 card is chosen at random
For 1 diamond card, probability = 13/52
For 1 queen, probability = 4/52
For cards which are both diamond and queen,
probability = 1/52
So required probability = 13/52 + 4/52 – 1/52 =
16/52 = 4/13
Question 16 of 25
16. Question
1 points
From a pack of 52 cards, 1 card is drawn at random. What is the probability of the card
being red or ace?
Correct
In 52 cards, there are 26 red cards and 4 ace and
there 2 such cards which are both red and ace.
1 card is chosen at random
For 1 red card, probability = 26/52
For 1 ace, probability = 4/52
For cards which are both red and ace, probability = 2/52
So required probability = 26/52 + 4/52 – 2/52 = 28/52 = 7/13
Incorrect
In 52 cards, there are 26 red cards and 4 ace and
there 2 such cards which are both red and ace.
1 card is chosen at random
For 1 red card, probability = 26/52
For 1 ace, probability = 4/52
For cards which are both red and ace, probability = 2/52
So required probability = 26/52 + 4/52 – 2/52 = 28/52 = 7/13
Question 17 of 25
17. Question
1 points
There are 250 tickets in an urn numbered 1 to 250. One ticket is chosen at random. What is the probability of it being a number containing a multiple of 3 or 8?
Correct
Multiples of 3 up to 250 = 250/3 = 83 (take only
whole number before the decimal part)
Multiples of 8 up to 250 = 250/3 = 31
Multiples of 24 (3×8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = 104/250 = 52/125
Incorrect
Multiples of 3 up to 250 = 250/3 = 83 (take only
whole number before the decimal part)
Multiples of 8 up to 250 = 250/3 = 31
Multiples of 24 (3×8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = 104/250 = 52/125
Question 18 of 25
18. Question
1 points
There are 4 white balls, 5 blue balls and 3 green balls in a box. 2 balls are chosen at
random. What is the probability of both balls being non-blue?
Correct
Both balls being non-blue means both balls are
either white or green
There are total 12 balls (4+3+5)
and total 7 white + green balls.
So required probability = 7C2/12C2 = [(7*6/2*1) /
(12*11/2*1)] = 21/66 = 7/22
Incorrect
Both balls being non-blue means both balls are
either white or green
There are total 12 balls (4+3+5)
and total 7 white + green balls.
So required probability = 7C2/12C2 = [(7*6/2*1) /
(12*11/2*1)] = 21/66 = 7/22
Question 19 of 25
19. Question
1 points
There are 4 white balls, 3 blue balls and 5 green balls in a box. 2 balls are chosen at random. What is the probability that first ball is green and second ball is white or green in color?
Correct
There are total 4+3+5 = 12 balls
Probability of first ball being green is = 5/12
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or
green is 8/12 = 2/3
So required probability = 5/12 * 2/3 = 5/18
Incorrect
There are total 4+3+5 = 12 balls
Probability of first ball being green is = 5/12
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or
green is 8/12 = 2/3
So required probability = 5/12 * 2/3 = 5/18
Question 20 of 25
20. Question
1 points
2 dices are thrown. What is the probability that there is a total of 7 on the dices?
Correct
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For a total of 7 on dices, we have – {1,6}, {6,1},
{2,5}, {5,2}, {3,4}, {4,3} – so 6 choices
So required probability = 6/36 = 1/6
Incorrect
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For a total of 7 on dices, we have – {1,6}, {6,1},
{2,5}, {5,2}, {3,4}, {4,3} – so 6 choices
So required probability = 6/36 = 1/6
Question 21 of 25
21. Question
1 points
2 dices are thrown. What is the probability that sum of numbers on the two dices is a
multiple of 5?
Correct
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For sum of number to be a multiple of 5, we
have – {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4},
{5,5} – so 7 choices
So required probability = 7/36
Incorrect
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For sum of number to be a multiple of 5, we
have – {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4},
{5,5} – so 7 choices
So required probability = 7/36
Question 22 of 25
22. Question
1 points
There are 25 tickets in a box numbered 1 to 25. 2 tickets are drawn at random. What is the probability of the first ticket being a multiple of 5 and second ticket being a multiple of 3.
Correct
There are 5 tickets which contain a multiple of 5
So probability of 1st ticket containing multiple of 5 = 5/25 = 1/5
Now:
Case 1: If the ticket chosen contained 15
If there was a 15 on first draw, then there are 7
tickets in box which contain a multiple of 3 out
of 24 tickets. (25/3 – 1 = 8 – 1 = 7) – because 15
is already out from the box
So probability = 7/24 (24 tickets remaining after 1st draw)
Case 2: If the ticket chosen contained other than
15 (5 or 10 or 20 or 25)
If 15 was not there on first draw, then there are 8
tickets in box which contain a multiple of 3 out
of 24 tickets. (25/3 = 8) – because 15 is already out from the box
So probability = 8/24 (24 tickets remaining after 1st draw)
Add the cases for probability of multiple of 3 on
second ticket, so prob. = 7/24 + 8/24 = 15/24
(added the cases because we want one of these cases to happen and not both)
So required probability = 1/5 * 15/24 = 1/8
(multiplied the cases because we want both to happen)
Incorrect
There are 5 tickets which contain a multiple of 5
So probability of 1st ticket containing multiple of 5 = 5/25 = 1/5
Now:
Case 1: If the ticket chosen contained 15
If there was a 15 on first draw, then there are 7
tickets in box which contain a multiple of 3 out
of 24 tickets. (25/3 – 1 = 8 – 1 = 7) – because 15
is already out from the box
So probability = 7/24 (24 tickets remaining after 1st draw)
Case 2: If the ticket chosen contained other than
15 (5 or 10 or 20 or 25)
If 15 was not there on first draw, then there are 8
tickets in box which contain a multiple of 3 out
of 24 tickets. (25/3 = 8) – because 15 is already out from the box
So probability = 8/24 (24 tickets remaining after 1st draw)
Add the cases for probability of multiple of 3 on
second ticket, so prob. = 7/24 + 8/24 = 15/24
(added the cases because we want one of these cases to happen and not both)
So required probability = 1/5 * 15/24 = 1/8
(multiplied the cases because we want both to happen)
Question 23 of 25
23. Question
1 points
What is the probability of selecting a two digit number at random such that it is a multiple
of 2 but not a multiple of 14?
Correct
There are 90 two digit numbers (10-99)
Multiple of 2 = 90/2 = 45
Multiple of 14 = 90/14 = 6
Since all multiples of 14 are also multiple of 2,
so favorable events = 45 – 6 = 39
So required probability = 39/90 = 13/30
Incorrect
There are 90 two digit numbers (10-99)
Multiple of 2 = 90/2 = 45
Multiple of 14 = 90/14 = 6
Since all multiples of 14 are also multiple of 2,
so favorable events = 45 – 6 = 39
So required probability = 39/90 = 13/30
Question 24 of 25
24. Question
1 points
There are 2 urns. 1st urn contains 6 white and 6 blue balls. 2nd urn contains 5 white and 7 black balls. One ball is taken at random from first urn and put to second urn without noticing its color. Now a ball is chosen at random from 2nd urn. What is the probability of the second ball being a white colored ball?
Correct
Case 1: first was a white ball
Now it is put in second urn, so total white balls
in second urn = 5+1 = 6, and total balls in
second urn = 12+1 = 13
So probability of white ball from second urn = 6/13
Case 2: first was a blue ball
Now it is put in second urn, so total white balls
in second urn remain 5, and total balls in second
urn = 12+1 = 13
So probability of white ball from second urn =5/13
So required probability = 6/13 + 5/13 = 11/13
(added the cases because we want one of these cases to happen and not both)
Incorrect
Case 1: first was a white ball
Now it is put in second urn, so total white balls
in second urn = 5+1 = 6, and total balls in
second urn = 12+1 = 13
So probability of white ball from second urn = 6/13
Case 2: first was a blue ball
Now it is put in second urn, so total white balls
in second urn remain 5, and total balls in second
urn = 12+1 = 13
So probability of white ball from second urn =5/13
So required probability = 6/13 + 5/13 = 11/13
(added the cases because we want one of these cases to happen and not both)
Question 25 of 25
25. Question
1 points
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards
are drawn and are found to be both hearts. Find the Probability of the lost card being a
heart?
Correct
Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50
Incorrect
Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50
