Probability Questions and Answers : Quantitative Aptitude

    Probability is defined as the chance that an event will happen or the likelihood that an event will happen.

    The definition of probability is


  • A family has two children. find the probability that both the children are girls given that at least

    one of them is a girl?

    A.) 1/4 B.) 2/3
    C.) 1/3 D.) 1/4
    Answer: Option 'C'

    Let b stand for boy and g for girl. The sample space of the experiment is

    S = {(g, g), (g, b), (b, g), (b, b)}

    Let E and F denote the following events :

    E : ‘both the children are girls’

    F : ‘at least one of the child is a girl’

    Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}

    Now E n F = {(g,g)}

    Thus P(F) = 3/4

    and P (E n F )= 1/4

    Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3


  • Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn

    randomly. If it is known that the number on the drawn card is more than 3, what is the probability

    that it is an even number?

    A.) 2/7 B.) 6/7
    C.) 7/2 D.) 4/7
    Answer: Option 'D'

    Let A be the event ‘the number on the card drawn is even’ and B be the

    event ‘the number on the card drawn is greater than 3’. We have to find P(A|B).

    Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

    Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10}

    and A n B = {4, 6, 8, 10}

    Also P(A) = 5/2, P(B) = 7/10 and P(A n B) = 4/10

    Then P(A|B) = P(A n B)/P(B) = (4/10)/(7/10) = 4/7


  • A die is thrown three times. Events X and Y are defined as below:

    X : 4 on the third throw

    Y : 6 on the first and 5 on the second throw

    What is the probability of X given that Y has already occurred.

    A.) 2/6 B.) 1/6
    C.) 1/5 D.) None of these
    Answer: Option 'B'

    The sample space has 216 outcomes.
    Now X =
    (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4)
    (3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4)
    (5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,5,4) (6,6,4)
    Y = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}

    and X n Y = {(6,5,4)}.

    Now P(Y) = 6/216

    and P (X n Y) = 1/216

    Then P(X|Y) = (1/216)/(6/216) = 1/6.


  • A die is thrown twice and the sum of the numbers appearing is observed to be 6.

    find the conditional probability that the number 4 has appeared at least once?

    A.) 1/5 B.) 3/5
    C.) 2/5 D.) None of these
    Answer: Option 'C'

    Let E be the event that ‘number 4 appears at least once’ and F be

    the event

    that ‘the sum of the numbers appearing is 6’.

    Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4),

    (5,4), (6,4)}

    and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}

    We have P(E) = 11/36

    and P (F) = 5/36

    Also EnF = {(2,4), (4,2)}

    Therefore P(EnF) = 2/36

    the required probability

    P(E|F) = P(EnF)/P(F) = (2/36)/(5/36) = 2/5.


  • Given that E and F are events such that P(E) = 0.16, P(F) = 0.4 and P(E n F) = 0.4,

    find P (E|F) and P(F|E)

    A.) 3/4,2 B.) 1/4,1
    C.) 1,1/4 D.) None of these
    Answer: Option 'C'

    Here, E and F are events

    P(E|F) = P(EnF)/P(F) = 0.4/0.4 = 1

    P(F|E) = P(EnF)/P(E) = 0.4/0.16 = 1/4.



Quantitative Aptitude Topics

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