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04 Jul 2019, 08:47
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A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?
A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)
A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)
04 Jul 2019, 08:47
Kudos
Bookmarks
Official Solution:
A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?
A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)
A die has two numbers greater than 4: 5 and 6. The probability of rolling a number greater than 4 (either 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\). Consequently, the probability of rolling a 4 or less is \(1- \frac{1}{3}= \frac{2}{3}\).
We want to determine the probability that a number greater than 4 will first appear on the third or fourth roll.
The probability of a number greater than 4 first appearing on the third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)
The probability of a number greater than 4 first appearing on the fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)
Therefore, the combined probability that a number greater than 4 will first appear on either the third or fourth roll is \(\frac{4}{27} + \frac{8}{81} = \frac{20}{81}\).
Answer: E
A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?
A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)
A die has two numbers greater than 4: 5 and 6. The probability of rolling a number greater than 4 (either 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\). Consequently, the probability of rolling a 4 or less is \(1- \frac{1}{3}= \frac{2}{3}\).
We want to determine the probability that a number greater than 4 will first appear on the third or fourth roll.
The probability of a number greater than 4 first appearing on the third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)
The probability of a number greater than 4 first appearing on the fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)
Therefore, the combined probability that a number greater than 4 will first appear on either the third or fourth roll is \(\frac{4}{27} + \frac{8}{81} = \frac{20}{81}\).
Answer: E
05 Jul 2019, 04:27
Kudos
Bookmarks
There are two cases to be considered to calculate the total probability in this question.
Case 1: Number greater than 4 appears on the 3rd throw &
Case 2: Number greater than 4 appears on the 4th throw.
When a fair dice is rolled, two numbers, which are greater than 4, can possibly come up viz 5 OR 6. Therefore, the other four numbers – 1,2 3 and 4 – are not greater than 4.
Therefore,
Probability of a number greater than 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\) &
Probability of a number not greater than 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\).
Case 1:
Number greater than 4 appears on the 3rd throw for the first time. This means, the first two throws showed numbers which were not greater than 4.
So, probability = \(\frac{2}{3} * \frac{2}{3} * \frac{1}{3}\) = \(\frac{4}{27}\).
Case 2: Number greater than 4 appears on the 4th throw for the first time. This means, the first three throws showed numbers which were not greater than 4.
Thus, probability = \(\frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{1}{3}\) = \(\frac{8}{81}\).
Therefore, total probability = \(\frac{4}{27}\) + \(\frac{8}{81}\)= \(\frac{20}{81}\).
The correct answer option is E.
In questions like these, it is very important to visualize and write down the cases, based on the constraints defined in the question. When you do this, you will be able to figure out the possible cases and once you get the cases, calculating probability is fairly easy.
Hope this helps!
Case 1: Number greater than 4 appears on the 3rd throw &
Case 2: Number greater than 4 appears on the 4th throw.
When a fair dice is rolled, two numbers, which are greater than 4, can possibly come up viz 5 OR 6. Therefore, the other four numbers – 1,2 3 and 4 – are not greater than 4.
Therefore,
Probability of a number greater than 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\) &
Probability of a number not greater than 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\).
Case 1:
Number greater than 4 appears on the 3rd throw for the first time. This means, the first two throws showed numbers which were not greater than 4.
So, probability = \(\frac{2}{3} * \frac{2}{3} * \frac{1}{3}\) = \(\frac{4}{27}\).
Case 2: Number greater than 4 appears on the 4th throw for the first time. This means, the first three throws showed numbers which were not greater than 4.
Thus, probability = \(\frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{1}{3}\) = \(\frac{8}{81}\).
Therefore, total probability = \(\frac{4}{27}\) + \(\frac{8}{81}\)= \(\frac{20}{81}\).
The correct answer option is E.
In questions like these, it is very important to visualize and write down the cases, based on the constraints defined in the question. When you do this, you will be able to figure out the possible cases and once you get the cases, calculating probability is fairly easy.
Hope this helps!
