Re: An n-sided die has sides labeled with the numbers 1 through n, inclusi
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09 Nov 2020, 12:16
A is the correct answer.
From statement 1)
This statement is telling us the dice has exactly 5 faces, since there is no other possibility that two rolls show two different faces with probability 80%!
The probability that the two rolls are different is --> (5/5) * (4/5) = 4/5 = 0,8.
Now, the probability that the two rolls are NOT different is 1-0,8 = 1/5 = 0,2.
PAY ATTENTION: this is not the probability to get 1 twice, but the probability that ANY of number of the dice is rolled twice.
Since the probability to get one number is distributed EQUALLY on the n=5 faces, the probability to roll any of the 5 numbers twice is (1/5)/(5) = 0,04 = 4% which is exactly what we are looking for!
OR
Since we know by the initial observation that the dice has n=5 faces, we can calculate the probability that one of the five faces (#1) will be rolled twice:
(1/5) * (1/5) = 1/25 = 4/100 = 4% which is exactly what we are looking for!!
SUFFICIENT
From statement 2)
Let's reason ad absurdum:
if the dice only had 1 face we would always get number 1, so the probability to get 1 twice is 100%.
If it had 2 faces, that is 2 numbers (1 and 2), the probability to get 1 twice would become 1/2 * 1/2= 1/4.
As we increase the number of faces, we notice that the probability to get the same number twice only decreases (3 faces --> 1/9, 4 faces --> 1/16).
Since we are told the probability to get the same number twice is 7/8, which is greater than 1/4, we can say this is nonsense and the information is clearly NOT sufficient.
Hope this helps you,
kudos are very welcome!
Jenp