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Probability Test 11
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Probability Test 11
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  • Question 1/19
    1 / -0

    A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?

    Solutions

    Explanation :

    Balls are picked in two manners – RWRW or WRWR

    So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99

     

  • Question 2/19
    1 / -0

    P and Q are sitting in a ring with 11 other persons. If the arrangement of 11 persons is at random, then the probability that there are exactly 4 persons between them?

    Solutions

    Explanation :

    Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12

    Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so favourable outcomes = 2

    So, probability = 2/12 = 1/6

     

  • Question 3/19
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    10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?

    Solutions

    Explanation :

    Total outcomes = (10 -1)! = 9!

    Favourable outcomes = 6!*4! (4 person seated together and 6 other persons seated randomly, so they will sit in (7-1)! Ways and those 4 persons can be arranged

    in 4! ways)

    So probability = 1/21

     

  • Question 4/19
    1 / -0

    A box contains 4 red, 5 black and 6 green balls. 3 balls are drawn at random. What is the probability that all the balls are of same colour?

    Solutions

    Explanation :

    (4c3 + 5c3 + 6c3)/15c3 = 34/455

     

  • Question 5/19
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    An apartment has 8 floors. An elevator starts with 4 passengers and stops at 8 floors of the apartment. What is the probability that all passengers travels to different floors?

    Solutions

    Explanation :

    Total outcomes = 8*8*8*8

    Favourable outcomes = 8*7*6*5 (first person having 8 choices, after that second person have 7 choices and so on)

    So, probability = 105/256

     

  • Question 6/19
    1 / -0

    A speak truth in 60% cases and B in 80% cases. In what percent of cases they likely to contradict each other narrating the same incident?

    Solutions

    Explanation :

    P(A) = 3/5 and P(B) = 4/5. Now they are contradicting means one is telling truth and other telling the lie. So,

    Probability = (3/5)*(1/5) + (2/5)*(4/5)

     

  • Question 7/19
    1 / -0

    A box contains 30 electric bulbs, out of which 8 are defective. Four bulbs are chosen at random from this box. Find the probability that at least one of them is defective?

    Solutions

    Explanation :

    1 – 22c4/30c4 = 1 – 209/783 = 574/783

     

  • Question 8/19
    1 / -0

    Two person A and B appear in an interview.  The probability of A’s selection is 1/5 and the probability of B’s selection is 2/7. What is the probability that only one of them is selected?

    Solutions

    Explanation :

    A selects and B rejects + B selects and A rejects = (1/5)*(5/7) + (4/5)*(2/7) = 13/35

     

  • Question 9/19
    1 / -0

    A 4- digit number is formed by the digits 0, 1, 2, 5 and 8 without repetition. Find the probability that the number is divisible by 5.

    Solutions

    Explanation :

    Total possibility = 5*4*3*2

    Favourable outcomes = 2*4*3*2  (to be divisible by 5 unit digit can be filled with only 0 or 5, so only two possibilities are there,  then the remaining can be filled in

    4, 3 and 2 ways respectively)

    so probability = 2/5

     

  • Question 10/19
    1 / -0

    A bag contains 6 red balls and 8 green balls. 2 balls are drawn at random one by one with replacement. Find the probability that both the balls are green:

    Solutions

    Explanation :

    (8c1)/(14c1) * (8c1)*(14c1) = 16/49

     

  • Question 11/19
    1 / -0

    A box contains 27 marbles some are blue and others are green. If a marble is drawn at random from the box, the probability that it is blue is 1/3. Then how many number of green marbles in the box?

    Solutions

    Explanation :

    Blue marble – x

    xc1/27c1 = 1/3

    x/27=1/3 —> x=27/3=9

    No of green marbles = Total – Blue marble =27-9=18

     

  • Question 12/19
    1 / -0

    In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random.What is the probability that at least one ball is of red colour?

    Solutions

    Explanation :

    Total Balls =10

    Other than red ball = 6c2

    6c2/10c2=1/3 —> 1-1/3 = 2/3

     

  • Question 13/19
    1 / -0

    Sahil has two bags (A & B) that contain green and blue balls.In the Bag ‘A’ there are 6 green and 8 blue balls and in the Bag ‘B’ there are 6 green and 6 blue balls. One ball is drawn out from any of these two bags. What is the probability that the ball drawn is blue?

    Solutions

    Explanation :

    Total balls in A bag = 14, Total balls in A bag = 12

    A bag = 1/2(8c1/14c1) = 2/7

    B bag = 1/2(6c1/12c1) = 1/4 —> total Probability = 2/7 + 1/4 =15/28

     

  • Question 14/19
    1 / -0

    In an examination, there are three sections namely Reasoning, Maths and English. Reasoning part contains 4 questions. There are 5 questions in maths section and 6 questions in English section. If three questions are selected randomly from the list of questions then what is the probability that all of them are from maths?

    Solutions

    Explanation :

    Total no of questions= 15

    Probability = 5c3/15c3 = 2/91

     

  • Question 15/19
    1 / -0

    A basket contains 5 red 4 blue 3 green marbles. If three marbles picked up random, What is the probability that either all are green or all are red?

    Solutions

    Explanation :

    Total Marbles = 12

    Either all are green or all are red = 5c3 + 3c3

    probability = 5c3 + 3c3/12c3 = 11/220 = 1/20

     

  • Question 16/19
    1 / -0

    A basket contains 5 red 4 blue 3 green marbles. If three marbles picked up random, What is the probability that at least one is blue?

    Solutions

    Explanation :

    Total Marbles = 12

    other than blue 8c3 / 12c3 = 14/55

    probability = 1-14/55 = 41/55

     

  • Question 17/19
    1 / -0

    A basket contains 5 red 4 blue 3 green marbles. If two marbles picked up random, What is the probability that both are red?

    Solutions

    Explanation :

    Total Marbles = 12

    Probability = 5c2 / 12c2 = 5/33

     

  • Question 18/19
    1 / -0

    A bag contains 5 red caps, 4 blue caps, 3 yellow caps and 2 green caps.If three caps are picked at random, what is the probability that two are red and one is green?

    Solutions

    Explanation :

    Total caps = 14

    Probability = 5c2 * 2c1/ 14c3 = 5/91

     

  • Question 19/19
    1 / -0

    A bag contains 2 red caps, 4 blue caps, 3 yellow caps and 5 green caps. If three caps are picked at random, what is the probability that none is green?

    Solutions

    Explanation :

    Total caps = 14

    Probability = 5c0 * 9c3/ 14c3 = 3/13

     

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