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25 July, 05:36

Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives

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  1. 25 July, 07:39
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    0.2946

    Step-by-step explanation:

    Number of tosses, n = 200

    P (obtaining a 5), p = 1/6

    q = 1 - p = 5/6

    Normal approximation for binomial distribution

    P (X < A) = P (Z < (A - mean) / standard deviation)

    Mean = np

    = 200 x 1/6

    = 33.33

    Standard deviation = √npq

    = √ (200 (1/6) (5/6))

    = 5.27

    P (at most 30 fives) = P (X ≤ 30)

    = P (Z < (30.5 - 33.33) / 5.27) (continuity correction of 0.5 is added to 30)

    = P (Z < - 0.54)

    = 0.2946
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