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26 May 2022, 01:30
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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)!
Difficulty:
95%
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Question Stats:
38% (02:23) correct
63%
(02:10)
wrong
based on 96
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A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?
A. 1/15
B. 1/10
C. 1/6
D. 1/5
E. 1/4
Are You Up For the Challenge: 700 Level Questions
A. 1/15
B. 1/10
C. 1/6
D. 1/5
E. 1/4
Are You Up For the Challenge: 700 Level Questions
26 May 2022, 03:15
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The only way that 3 draws are needed is if the first two numbers drawn are either: \(1\)&\(2\), \(2\)&\(1\), \(1\)&\(3\) or \(3\)&\(1\) (4 pairs)
Number of ways in which two numbers can be drawn from 5: \(2*\frac{5!}{2!3!}= 20\)
Which means that the probability of there being 3 draws is: \(\frac{4}{20}\) or \(\frac{1}{5}\)
Answer D
Number of ways in which two numbers can be drawn from 5: \(2*\frac{5!}{2!3!}= 20\)
Which means that the probability of there being 3 draws is: \(\frac{4}{20}\) or \(\frac{1}{5}\)
Answer D
26 May 2022, 05:07
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In order to have to draw 3 times before the sum of the value exceeds 4.
The first draw cannot be 5, as it will end the sequence.
The first draw cannot be 4, as the sequence will end after the second draw.
The first draw can be 3, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20
The first draw can be 2, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20
The first draw can be 1, then the second draw must be either 2 or 3, (and the third draw can be anything).
Probability for this sequence is 1/5 * 2/4 * 3/3 = 2/20
Altogether, the probability is 1/20 + 1/20 + 2/20 = 4/20 = 1/5
The answer is thus (D).
The first draw cannot be 5, as it will end the sequence.
The first draw cannot be 4, as the sequence will end after the second draw.
The first draw can be 3, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20
The first draw can be 2, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20
The first draw can be 1, then the second draw must be either 2 or 3, (and the third draw can be anything).
Probability for this sequence is 1/5 * 2/4 * 3/3 = 2/20
Altogether, the probability is 1/20 + 1/20 + 2/20 = 4/20 = 1/5
The answer is thus (D).
