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What is the probability of rolling three fair dice and having at least
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01 Oct 2021, 09:34
01 Oct 2021, 09:34
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What is the probability of rolling three fair dice and having at least one of the three dice show an even number?
A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
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Re: What is the probability of rolling three fair dice and having at least
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01 Oct 2021, 09:51
01 Oct 2021, 09:51
2
Carcass wrote:
What is the probability of rolling three fair dice and having at least one of the three dice show an even number?
A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
We want P(select at least 1 even number)
When it comes to probability questions involving "at least," it's usually best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 even number) = 1 - P(not getting at least 1 even number)
What does it mean to not get at least 1 even number?
It means getting zero even numbers.
So, we can write: P(getting at least 1 even number) = 1 - P(getting zero even numbers)
P(getting zero even numbers) = P(1st coin is odd AND 2nd coin is odd AND 3rd coin is odd )
= P(1st coin is odd) x P(2nd coin is odd) x P(3rd coin is odd )
= 1/2 x 1/2 x 1/2
= 1/8
So, P(getting at least 1 even number) = 1 - 1/8
= 7/8
Answer: B
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Re: What is the probability of rolling three fair dice and having at least
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14 Oct 2022, 09:42
14 Oct 2022, 09:42
1
We need to find What is the probability of rolling three fair dice and having at least one of the three dice show an even number?
As we are rolling three dice => Number of cases = \(6^3\) = 216
Let's solve the problem using two methods:
Method 1:
Now there are 8 outcomes possible
(Odd, Odd, Odd), (Odd, Odd, Even), (Odd, Even Odd), (Odd, Even, Even), (Even, Odd, Odd), (Even, Odd, Even), (Even, Even Odd), (Even, Even, Even) and there is an equal chance of each of them happening
=> P(At least one number is Even) = \(\frac{7}{8}\) (all cases except Odd, Odd, Odd)
Method 2:
Out of the 216 comes lets eliminate all options in which all three outcomes are odd
All three can be odd in 3*3*3 (=27 ways), as in each roll we can get any number out of 1, 3, and 5 => 3 choices in each roll
=> P(At least one number is Even) = \(\frac{216 - 27}{216}\) = \(\frac{189}{216}\) = \(\frac{7}{8}\)
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 three dice => Number of cases = \(6^3\) = 216
Let's solve the problem using two methods:
Method 1:
Now there are 8 outcomes possible
(Odd, Odd, Odd), (Odd, Odd, Even), (Odd, Even Odd), (Odd, Even, Even), (Even, Odd, Odd), (Even, Odd, Even), (Even, Even Odd), (Even, Even, Even) and there is an equal chance of each of them happening
=> P(At least one number is Even) = \(\frac{7}{8}\) (all cases except Odd, Odd, Odd)
Method 2:
Out of the 216 comes lets eliminate all options in which all three outcomes are odd
All three can be odd in 3*3*3 (=27 ways), as in each roll we can get any number out of 1, 3, and 5 => 3 choices in each roll
=> P(At least one number is Even) = \(\frac{216 - 27}{216}\) = \(\frac{189}{216}\) = \(\frac{7}{8}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
