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AnilSinghBora
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A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment] Q.24

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I don’t know how to solve this problem please help me for solving the problem of question no.24 of probability chapter of class 10th ncert math how i solve this question in simple way A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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  1. Outcomes are:

    (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

    (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

    (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

    (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

    (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

    (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

    So, the total number of outcome = 6×6 = 36

    (i) Method 1:

    Consider the following events.

    A = 5 comes in first throw,

    B = 5 comes in second throw

    P(A) = 6/36,

    P(B) = 6/36 and

    P(not B) = 5/6

    So, P(not A) = 1-(6/36) = 5/6

    ∴ The required probability = (5/6)×(5/6) = 25/36

    Method 2:

    Let E be the event in which 5 does not come up either time.

    So, the favourable outcomes are [36–(5+6)] = 25

    ∴ P(E) = 25/36

    (ii) Number of events when 5 comes at least once = 11(5+6)

    ∴ The required probability = 11/36

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