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07 Jun 2018, 04:03
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:55) correct
30%
(02:19)
wrong
based on 220
sessions
History
Date
Time
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Not Attempted Yet
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled 4 times. What is the probability that on at least one roll the number showing will be less than 3?
A) \(\frac{65}{81}\)
B) \(\frac{67}{81}\)
C) \(\frac{8}{9}\)
D) \(\frac{26}{27}\)
E) \(\frac{80}{81}\)
A) \(\frac{65}{81}\)
B) \(\frac{67}{81}\)
C) \(\frac{8}{9}\)
D) \(\frac{26}{27}\)
E) \(\frac{80}{81}\)
07 Jun 2018, 04:18
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Solution
Given:
- • A fair die is rolled 4 times
To find:
- • The probability of showing at least one roll less than 3
Approach and Working:
- • The probability of showing at least one roll less than 3 = 1 – the probability of showing all greater than or equal to 3
= \(1 – (\frac{4}{6} * \frac{4}{6} * \frac{4}{6} * \frac{4}{6}) = 1 – \frac{16}{81} = \frac{65}{81}\)
Hence, the correct answer is option A.
Answer: A
07 Jun 2018, 04:27
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CAMANISHPARMAR
Probability that an outcome less than 3 occurs on one roll= \(\frac{2}{6}\)
--> probability that an outcome less than 3 does not occurs on one roll= 1- \(\frac{2}{6}\) = \(\frac{4}{6}\)
Now, probability that an outcome less than 3 occurs on at least one roll= 1- (probability that an outcome of less than 3 occurs on none of the 4 rolls)
=1- (\(\frac{4}{6}\)*\(\frac{4}{6}\)* \(\frac{4}{6}\)*\(\frac{4}{6}\)) = 1-\((\frac{16}{81})\)=\(\frac{65}{81}\)
Hence A is the answer.
