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Updated on: 22 Oct 2017, 20:37
Originally posted by Rorschach1337 on 22 Oct 2017, 20:11.
Last edited by Bunuel on 22 Oct 2017, 20:37, edited 1 time in total.
Last edited by Bunuel on 22 Oct 2017, 20:37, edited 1 time in total.
Renamed the topic.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4
25 Mar 2019, 02:34
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Rorschach1337
Total rolls = 4 times
Dice = 6 sided
Probability of a "2" on every roll = 1/6
Probability of not a "2" but any of the other 5 numbers = 1 - 1/6 = 5/6
Probability of "2" at least 3 times means :
a. 2 three times
b. 2 four times
Case 1 : T T T N (T = two and N = Not two)
\(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{5}{6} * \frac{4!}{3!}\)
We multiply by 4! because permutations/total arrangements of 4 different items can be done in 4! ways
We divide by 3! because we have 3 identical items.
Therefore, 4*(1/6)^4 * 5/6
Case 2 : T T T T (All twos)
\((1/6)^4\)
Add Case I and II
\(4*(1/6)^4 * 5/6\) + \((1/6)^4\)
General Discussion
22 Oct 2017, 20:34
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Rorschach1337
ATLEAST 3 times means 3 times and 4 times..
1) 3 times :-
3 times "2" will mean a probability of \(\frac{1}{6}^3\) and prob of any other number in 4 throw is 5/6..
but this ANY other number can come in any of 4 throws ..
so prob of 3times = \(\frac{1}{6}^3*4*\frac{5}{6}\)
2) 4 times :-
\(\frac{1}{6}^4\)
total \(\frac{1}{6}^4+4*\frac{1}{6}^3\frac{5}{6}\)
D
