- 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.
Probability of choosing a girl = 3/5 x/(14+x) = 3/5
=> x = 21
Required Probability = [14C1 * 21C1 + 14C2]/(14+21)C2
- 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.
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
- 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.
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
- 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?
Probability that atleast one bulb is defective = 1 – P (All are non-defective)
= 1 – 15C3/20C3 = 137/228
- 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.
Total number of balls in the bag = x+x+2+x+5
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
- 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?
Required Probabilty = [(6C2*5C1*1C1) + (6C2*5C2)]/12C4
= [(15*5*1)+(15*10)]/495 = 5/11
- 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.
1/(x+1) + 1/(x+3) + 1/5 = 59/120
=> 7x^2 – 20x – 75 = 0
=> 7x^2 – 35x + 15x – 75 = 0
=>x = 5
- 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.
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
- 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?
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
- 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?
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