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NDA - Probability Test 1186
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NDA - Probability Test 1186
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  • Question 1/5
    1 / -0.33
    In a multiple-choice test, an examinee either knows the correct answer with probability p or guesses with probability  The probability of answering a question correctly is  if he or she merely guesses. If the examinee answers a question correctly, the probability that he or she really knows the answer is
    Solutions

    Probability of knowing correct answer = p

    Probability to guess correct answer =

    Probability to answer correctly =

    So required probability =

  • Question 2/5
    1 / -0.33
    Five sticks of length 1, 3, 5, 7 and 9 feet are given. Three of these sticks are selected at random. What is the probability that the selected sticks can form a triangle?
    Solutions

    Five sticks of length 1, 3, 5, 7 and 9 feet are given

    3 sticks can be selected from 5 sticks in 5C3 ways

    5C3 =

    Probable cases = (3,5,7) , (5,7,9) and (3,7,9)

    (sum of the two smallest sides must be greater than the third one

    So

  • Question 3/5
    1 / -0.33
    Consider the following statements:

    1)

    2)

    3) P(A ∩ B) = P(B) P(A | B)

    Which of the above statements are correct?

    Solutions

    Statements 1 and 3 are correct.

     3) P(A | B)= P(A ∩ B)/P(B)
      
          P(A ∩ B) = P(B) P(A | B)

  • Question 4/5
    1 / -0.33
    Tickets numbered from 1 to 12 are mixed up together, and then a ticket is withdrawn at random. Find the probability that the ticket has a number which is a multiple of 2 or 3 .
    Solutions

    We know that,

    Probability of occurrence of an event =

     are 12

    Desired output is picking a number which is multiple of 2 or 3. So, desired outputs are 2, 3, 4, 6, 8, 9, 10, 12.

    Total no. of desired outputs are 8.

    Therefore, the Probability of getting a number which is multiple of 2 or 3 =

    =

  • Question 5/5
    1 / -0.33
    In class, 30% of the students offered mathematics 20% offered chemistry and 10% offered both. If a student is selected at random, find the probability that he has offered mathematics or chemistry.
    Solutions

    Given: Math students = 30%

    Chemistry students = 20%

    Math & Chemistry both = 10%

    To Find: P (Math or Chemistry)

    Now, P (Math) = 30% =  = 0.30

    P (Chemistry) = 20% =  = 0.20

    P (Math ∩ Chemistry) = 10% =  = 0.10

    We know that,

    P ( A ∪ B ) = P(A) + P(B) – P (A∩B)

    Therefore,

    P (Math ∩ Chemistry) = 0.30 + 0.20 – 0.10 = 0.40

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