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Two dice are tossed once. The probability of getting an even number at
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20 Sep 2021, 05:01
20 Sep 2021, 05:01
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Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: Two dice are tossed once. The probability of getting an even number at
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20 Sep 2021, 06:16
20 Sep 2021, 06:16
1
P(even)=\(1/2*1/2\)*\(2\), as two orders of dice can be set
P(total of 8)=\(1/6*1/6*5\), as \(4-4\) is counted once
P(even or total 8)=\(1/6*1/6*3\), as \(3-5\) and \(5-3\) pairs not counted
Required probability = P(even) + P(total of 8) - P(even or total 8)= \(1/2\) + \(5/36\) - \(3/36\) = \(18+5-3/36\) = \(20/36\)
Answer is D
P(total of 8)=\(1/6*1/6*5\), as \(4-4\) is counted once
P(even or total 8)=\(1/6*1/6*3\), as \(3-5\) and \(5-3\) pairs not counted
Required probability = P(even) + P(total of 8) - P(even or total 8)= \(1/2\) + \(5/36\) - \(3/36\) = \(18+5-3/36\) = \(20/36\)
Answer is D
GeminiHeat wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
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WE:Engineering (Computer Software)
Two dice are tossed once. The probability of getting an even number at
[#permalink]
16 Oct 2022, 09:47
16 Oct 2022, 09:47
Given that Two dice are tossed once and We need to find What is the The probability of getting an even number at the first die or a total of 8?
As we are rolling two dice => Number of cases = \(6^2\) = 36
Getting even number in the first dice
Possible cases for the two rolls _ _ are 3 * 6 = 18 ways
(As First dice can get even number in 3 ways (2, 4, 6) and second dice can get any of the 6 numbers
Getting a Sum of 8
We can get the sum of 8 in the following ways
(2, 6), (3, 5), (4, 4), (5, 3), (6, 6)
But all cases except (3, 5) and (5, 3) are considered above when we took first dice roll as an even number
=> 2 ways
=> Total ways = 18 + 2 = 20 ways
=> Probability of getting an even number at the first die or a total of 8 = \(\frac{20}{36}\)
So, Answer will be D
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
Getting even number in the first dice
Possible cases for the two rolls _ _ are 3 * 6 = 18 ways
(As First dice can get even number in 3 ways (2, 4, 6) and second dice can get any of the 6 numbers
Getting a Sum of 8
We can get the sum of 8 in the following ways
(2, 6), (3, 5), (4, 4), (5, 3), (6, 6)
But all cases except (3, 5) and (5, 3) are considered above when we took first dice roll as an even number
=> 2 ways
=> Total ways = 18 + 2 = 20 ways
=> Probability of getting an even number at the first die or a total of 8 = \(\frac{20}{36}\)
So, Answer will be D
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
