Timothy creates a game in which the player rolls 4 dice. What is the probability in this game of having exactly two dice or more land on a f

Question

Timothy creates a game in which the player rolls 4 dice. What is the probability in this game of having exactly two dice or more land on a five?
A. 0.016
B. 0.132
C. 0.868
D. 0.984

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Adeline 5 days 2021-10-14T13:23:47+00:00 1 Answer 0

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    2021-10-14T13:25:06+00:00

    Answer:

    B. 0.132

    Step-by-step explanation:

    For each time the dice is thrown, there are only two possible outcomes. Either it lands on a five, or it does not. The probability of a throw landing on a five is independent of other throws. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

    C_{n,x} = \frac{n!}{x!(n-x)!}

    And p is the probability of X happening.

    Timothy creates a game in which the player rolls 4 dice.

    This means that n = 4

    The dice can land in 6 numbers, one of which is 5.

    This means that p = \frac{1}{6}

    What is the probability in this game of having exactly two dice or more land on a five?

    P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

    In which

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 2) = C_{4,2}.(\frac{1}{6})^{2}.(\frac{5}{6})^{2} = 0.116

    P(X = 3) = C_{4,2}.(\frac{1}{6})^{3}.(\frac{5}{6})^{1} = 0.015

    P(X = 4) = C_{4,4}.(\frac{1}{6})^{4}.(\frac{5}{6})^{0} = 0.001

    P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.116 + 0.015 + 0.001 = 0.132

    So the correct answer is:

    B. 0.132

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45:7+7-4:2-5:5*4+35:2 =? ( )