# Quantitative Aptitude: Probability Questions – Set 9

1. In a group of 14 boys and x number of girls, the probability of choosing a girl is 3/5. If we have to select two students, find the probability that atleast one of them is boy.
12/17
5/9
10/13
11/17
13/15
Option D
Probability of choosing a girl = 3/5 x/(14+x) = 3/5
=> x = 21
Required Probability = [14C1 * 21C1 + 14C2]/(14+21)C2
= 11/17

2. A bag contains 8 red balls and x blue balls and the probability of choosing a blue ball is 3/5. If we randomly select two balls, find the probability that atleast one of them is red.
55/91
63/94
62/95
61/93
59/95
Option C
Probability of choosing a blue ball = 3/5
=> x/(x+8) = 3/5
=> x = 12
Required Probability = [8C1*12C1+8C2]/(8+12)C2 = 62/95

3. A bag contains 6 apples, 8 bananas and (x+2) oranges. Two fruits are chosen at random. Find the value of x if the probability that both fruits are oranges is 2/51.
2
5
3
4
6
Option A
Probability of selecting two oranges = (x+2)C2/(16+x)C2 Now, (x+2)C2/(16+x)C2 = 2/51
=> 7x^2 + 13x – 54 = 0
=> 7x^2 – 14x + 27x – 54 = 0
=> x = 2 or -27/7

4. A box contains 20 bulbs out of which 5 are defective. Three bulbs are randomly taken out of the box. What is the probability that out of the three at least one bulb is defective?
128/221
137/228
131/220
129/228
130/221
Option B
Probability that atleast one bulb is defective = 1 – P (All are non-defective)
= 1 – 15C3/20C3 = 137/228

5. A bag contains ‘x’ red balls , ‘x+2’ pink balls. Two balls are randomly drawn from the bag and the probability that a red and a blue ball are drawn is 4/21. Find the total number of balls in the bag.
33
40
38
49
30
Option D
Total number of balls in the bag = x+x+2+x+5
= 3x+7
Probability that a red and a blue ball are drawn = (xC1*(x+2)C1)/(3x+7)C2 = 4/21
2x(x+2)/(3x+7)(3x+6) = 4/21
=> x = 14
The number of balls in the bag = 14*3+7 = 49

6. In a bag there are 6 red balls, 5 white balls and 1 black ball. A man draws 4 balls at random from the bag. What will be the probability that 2 balls are red?
4/15
7/13
5/11
5/12
3/10
Option C
Required Probabilty = [(6C2*5C1*1C1) + (6C2*5C2)]/12C4
= [(15*5*1)+(15*10)]/495 = 5/11

7. In IPL 2010, the chances of team CSK winning are 1/(x+1), the chances of team KKR winning are 1/(x+3) and chances of winning of Mumbai Indian are 1/5. If total 8 teams are there and the probability of winning of one of these three teams (CSK, KKR and Mumbai Indians) is 59/120, find teh value of x.
4
7
6
8
5
Option E
1/(x+1) + 1/(x+3) + 1/5 = 59/120
=> 7x^2 – 20x – 75 = 0
=> 7x^2 – 35x + 15x – 75 = 0
=>x = 5

8. A bag contains red, blue and green balls in the ratio of 3:5:4 resp. 10 pink balls are put in the bag and two balls are randomly drawn from the bag. The probability that one ball is green and other is red is 20/161. Find the difference in the number of green and red balls in the bag.
6
5
7
4
2
Option B
Probability that one ball is red and other is green
= (3xC1*4xC1)/ (12x+10)C2 = 20/161
=> 246x^2 – 1140x – 450 = 0
=> 41x^2 – 190x – 75 = 0
=> x = 5, -15/41
Bag contains 10 pink, 15 red , 25 blue and 20 green balls.
Difference in the number of green and red balls = 20 – 15 = 5

9. There are ‘x’ bottles and ‘y ’ glasses in a tray and the probability of randomly picking a bottle is 2/5. Four bottles are added to the tray and the probability of picking a bottle becomes is 4/7. What was the number of glasses in th tray?
8
4
5
6
9
Option D
x/(x+y) = 2/5
=> 3x = 2y —-(1)
(x+4)/(x+y+4) = 4/7
=> 3x+12 = 4y —-(2)
On solving these two equations, we get
x = 4 and y = 6

10. A bag contains 22 roses of three different colours like yellow, white and pink. The ratio of yellow roses to pink roses is 1:2 resp. and the probability of choosing two white rose from the bag is 1/11. If two roses are picked from the bag. What is the probability that one rose is white and other one is pink?
13/30
10/33
15/37
10/29
11/31
Option B
Let the number of white roses be x.
Probability of choosing two white roses = 1/11
xC2/22C2 = 1/11
=> x = 7
Number of yellow and pink roses = 22 – 7 = 15
Number of pink roses = 2/(1+2)*15 = 10
Required Probability = 7C1*10C1/22C1 = 10/33