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

Question

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}

Princ · Accepted Answer

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}

DavidTutorexamPAL · Answer

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.

(D) is our answer.

sudarshan22 · Answer

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  .

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=5PVZYQXet18

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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

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