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A number cube has six faces numbered 1 through 6. If the cube is rolle

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A number cube has six faces numbered 1 through 6. If the cube is rolle  [#permalink]

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31 Jul 2018, 00:08
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65% (01:31) correct 35% (01:25) wrong based on 54 sessions

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A number cube has six faces numbered 1 through 6. 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?

(A) $$\frac{2}{9}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{4}{9}$$

(D) $$\frac{5}{9}$$

(E) $$\frac{2}{3}$$

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

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31 Jul 2018, 00:16
Bunuel wrote:
A number cube has six faces numbered 1 through 6. 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?

(A) $$\frac{2}{9}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{4}{9}$$

(D) $$\frac{5}{9}$$

(E) $$\frac{2}{3}$$

In questions asking us to calculate 'at least one of' something, it is usually easier to calculate the complement.
This is a Logical approach approach.

The complement of 'at least one roll greater than four' is 'none of the rolls are greater than four' which is the same as 'all of the rolls are smaller than 5'.
The probability to get a number smaller than 5 is 4/6 and the probability that both rolls are smaller than 5 is (4/6)^2 = (2/3)^2 = 4/9.
So our answer is 1 - 4/9 = 5/9.

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

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31 Jul 2018, 02:22
Bunuel wrote:
A number cube has six faces numbered 1 through 6. 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?

(A) $$\frac{2}{9}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{4}{9}$$

(D) $$\frac{5}{9}$$

(E) $$\frac{2}{3}$$

OA: D
The probability that at least one of the rolls will result in a number greater than 4 can be found out by taking these 3 cases

Case 1: {5,6} on 1st roll and {1,2,3,4} on 2nd roll
$$Probability_{case1} =\frac{2}{6}*\frac{4}{6}=\frac{8}{36}$$

Case 2: {1,2,3,4} on 1st roll and {5,6} on 2nd roll
$$Probability_{case2} =\frac{4}{6}*\frac{2}{6}=\frac{8}{36}$$

Case 3: {5,6} on 1st roll and {5,6} on 2nd roll
$$Probability_{case3} =\frac{2}{6}*\frac{2}{6}=\frac{4}{36}$$

$$Total Probability = Probability_{case1}+Probability_{case2}+Probability_{case3}$$
$$=\frac{8}{36}+\frac{8}{36}+\frac{4}{36}$$
$$=\frac{20}{36} =\frac{5}{9}$$
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Re: A number cube has six faces numbered 1 through 6. If the cube is rolle  [#permalink]

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31 Jul 2018, 02:43
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Bunuel wrote:
A number cube has six faces numbered 1 through 6. 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?

Possible value in one throw of the dice = { 1, 2, 3, 4, 5, 6 }
Values less than 5 = 4
Total possible values = 6

Probability of result less than 5 = Values less than 5 / Total possible values = 4 / 6
Since the dice is thrown twice = (4/6)^2 = 16/36 = 4/9

Probability of at least one of the rolls will result in a number greater than 4 = 1 - 4/9 = 5/9

Hence, D.
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Re: A number cube has six faces numbered 1 through 6. If the cube is rolle &nbs [#permalink] 31 Jul 2018, 02:43
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