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13 May 2024, 00:10
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:14) correct
40%
(02:57)
wrong
based on 177
sessions
History
Date
Time
Result
Not Attempted Yet
If four dice are rolled, what is the probability that at least two of Them show the same face?
A) 37/72
B) 11/36
C) 47/72
D) 25/36
E) 13/18
A) 37/72
B) 11/36
C) 47/72
D) 25/36
E) 13/18
01 Jun 2024, 00:01
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Total possible outcomes when rolling four dice:
Each die has 6 faces, so there are:
6^4=1296, possible outcomes when rolling four dice.
Calculate the number of outcomes where all dice show different faces:
The first die can show any of the 6 faces.
The second die can show any of the remaining 5 faces (different from the first die).
The third die can show any of the remaining 4 faces (different from the first two dice).
The fourth die can show any of the remaining 3 faces (different from the first three dice).
Therefore, the number of ways in which all four dice show different faces is:
6×5×4×3=360
Calculate the probability that all dice show different faces:The probability 𝑃(all different) divided by the total number of outcomes:
𝑃(all different)=360/ 1296=5/18
Calculate the probability that at least two dice show the same face:
𝑃(at least two the same)=1−𝑃(all different)=1−5/18=13/18
Each die has 6 faces, so there are:
6^4=1296, possible outcomes when rolling four dice.
Calculate the number of outcomes where all dice show different faces:
The first die can show any of the 6 faces.
The second die can show any of the remaining 5 faces (different from the first die).
The third die can show any of the remaining 4 faces (different from the first two dice).
The fourth die can show any of the remaining 3 faces (different from the first three dice).
Therefore, the number of ways in which all four dice show different faces is:
6×5×4×3=360
Calculate the probability that all dice show different faces:The probability 𝑃(all different) divided by the total number of outcomes:
𝑃(all different)=360/ 1296=5/18
Calculate the probability that at least two dice show the same face:
𝑃(at least two the same)=1−𝑃(all different)=1−5/18=13/18
General Discussion
13 May 2024, 00:14
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I tried solving the following way but didn’t get the correct answer:
1 - (None has the same face) = 1 - (ALL faces are different)
1 - (6C4)/(6^4) = 1 - (5/432)
Can anyone guide why this is wrong?
Also, why can’t we choose all the different faces as
6C1 * 5C1 * 4C1 * 3C1
But this is different from 6C4?
Do we need to arrange them as 4!
Should it be 6C4 * 4!
Please guide on how to select four different faces?
And how to solve by this approach?
1 - (None has the same face) = 1 - (ALL faces are different)
1 - (6C4)/(6^4) = 1 - (5/432)
Can anyone guide why this is wrong?
Also, why can’t we choose all the different faces as
6C1 * 5C1 * 4C1 * 3C1
But this is different from 6C4?
Do we need to arrange them as 4!
Should it be 6C4 * 4!
Please guide on how to select four different faces?
And how to solve by this approach?
