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Re: A number cube has six faces numbered 1 through 6. If the cub [#permalink]
Given that A number cube has six faces numbered 1 through 6 and We need to find If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4?

As we are rolling the cube twice => Number of cases = \(6^2\) = 36

P(at least one of the rolls will result in a number greater than 4) = 1 - P(None of the two rolls will result in a number greater than 4)

P(None of the two rolls will result in a number greater than 4) = P(Both the rolls will have a number from 1 to 4)

Following are the possible cases:
(1,1), (1,2), (1,3), (1,4)
(2,1), (2,2), (2,3), (2,4)
(3,1), (3,2), (3,3), (3,4)
(4,1), (4,2), (4,3), (4,4)

=> 16 cases

=> P(None of the two rolls will result in a number greater than 4) = \(\frac{16}{36}\) = \(\frac{4}{9}\)

=> P(at least one of the rolls will result in a number greater than 4) = 1 - P(None of the two rolls will result in a number greater than 4)
= 1 - \(\frac{4}{9}\)
= \(\frac{9 - 4}{9}\)
= \(\frac{5}{9}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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Re: A number cube has six faces numbered 1 through 6. If the cub [#permalink]
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