What is the probability of rolling two fair dice and having at least

Question

What is the probability of rolling two fair dice and having at least one die show an even number?

A. 35/36
B. 3/4
C. 1/2
D. 1/4
E. 1/36

vitaliyGMAT · Answer

P (at least one even) = 1 - P(all odd) =   1 - \frac{1}{2}*\frac{1}{2} = 1 - \frac{1}{4} = \frac{3}{4}

Answer B

kishblossom · Answer

Answer can be B. 3/4
1/2 + 1/4

Bhavanahiremath · Answer

Probability of first one being even and second one being odd number = 3/6*3/6
Probability of first one being odd and second being even = 3/6*3/6
Probability of both being even = 3/6*3/6
Total = 3/6*3/6+3/6*3/6+3/6*3/6 = 3/4 Ans.B

AnisMURR · Answer

P (at least one even) = 1 - P(all odd) = 1- 0.5*0.5= 0.75

So the Answer is B

anupam87 · Answer

P(success) = 1 - P(zero success) = 1 - (1/2*1/2) = 3/4

whenever the question says "at least" it's always a better approach to go for complement rule
P(success) = 1 - P(zero success)

Posted from my mobile device

Nunuboy1994 · Answer

In order to solve this question we need to know the probability of rolling an even number on each individual dice. Dices are numbered 1-6 so the probability of rolling an even number is 3/6 and 3/6. We then use the formula.

P (at least one even) = 1 - P(all odd) =

P (at least one even) = 36/36 - P(9/36) = 27/36

simplify

Thus 
B. 3/4

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=4zTFY2pUFbs

We need to find What is the probability of rolling two fair dice and having at least one die show an even number?

As we are rolling two dice => Number of cases =  6^2  = 36

Let's solve the problem using two methods:

Method 1:

Now there are 4 outcomes possible
(Odd, Odd), (Odd, Even), (Even Odd), (Even, Even) and there is an equal chance of each of them happening

=> P(At least one number is Even) = P((Odd, Even) or (Even Odd) or (Even, Even)) =  \frac{3}{4}

Method 2:

Out of the 36 comes lets eliminate all options in which both the outcomes are odd
Both can be odd in 3*3 (=9 ways), as in the first roll we can get any number out of 1, 3, and 5. And in the second case also we have these 3 choices.

=> P(At least one number is Even) =  \frac{36 - 9}{36}  =  \frac{27}{36}  =  \frac{3}{4}

So,  Answer will be B 
Hope it helps!

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

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

Gijmaja · Answer

no even number = 1/2*1/2=1/4

Prob of at least 1 = 1-1/4=3/4.

bumpbot · Answer

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