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What is the probability of rolling two fair dice and having
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13 Apr 2020, 09:37
13 Apr 2020, 09:37
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What is the probability of rolling two fair dice and having each die show an odd number?
A. 1/6
B. 1/4
C. 1/2
D. 3/4
E. 35/36
Kudos for the right answer and explanation
A. 1/6
B. 1/4
C. 1/2
D. 3/4
E. 35/36
Kudos for the right answer and explanation
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Joined: 10 Apr 2015
Posts: 6218
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Re: What is the probability of rolling two fair dice and having
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13 Apr 2020, 10:56
13 Apr 2020, 10:56
Carcass wrote:
What is the probability of rolling two fair dice and having each die show an odd number?
A. 1/6
B. 1/4
C. 1/2
D. 3/4
E. 35/36
Kudos for the right answer and explanation
A. 1/6
B. 1/4
C. 1/2
D. 3/4
E. 35/36
Kudos for the right answer and explanation
Each die has six possible outcomes: 1, 2, 3, 4, 5, or 6
Three of those outcomes are odd (1, 3 and 5).
So, P(a die is odd) = 3/6 = 1/2
P(both are odd) = P(1st die is odd AND 2nd die is odd)
= P(1st die is odd) x P(2nd die is odd)
= 1/2 x 1/2
= 1/4
Answer: B
Cheers,
Brent
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Re: What is the probability of rolling two fair dice and having
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15 Oct 2022, 10:25
15 Oct 2022, 10:25
We need to find What is the probability of rolling two fair dice and having each die show an odd 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(Both will be odd) = \(\frac{1}{4}\)
Method 2:
Both the die 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.
(1,1), (1,3), (1,5)
(3,1), (3,3), (3,5)
(5,1), (5,3), (5,5)
=> P(each die show an odd number) = \(\frac{9}{36}\) = \(\frac{1}{4}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
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(Both will be odd) = \(\frac{1}{4}\)
Method 2:
Both the die 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.
(1,1), (1,3), (1,5)
(3,1), (3,3), (3,5)
(5,1), (5,3), (5,5)
=> P(each die show an odd number) = \(\frac{9}{36}\) = \(\frac{1}{4}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
