Quantitative Aptitude: Probability Questions – Set 9

Banya
  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

     


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