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16 Sep 2014, 00:50
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D
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77% (01:54) correct
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In a basket, there are 3 white balls and 5 blue balls. If Mary draws one ball at random and keeps it, what is the probability that John will subsequently draw a blue ball?
A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)
A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)
16 Sep 2014, 00:50
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Official Solution:
In a basket, there are 3 white balls and 5 blue balls. If Mary draws one ball at random and keeps it, what is the probability that John will subsequently draw a blue ball?
A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)
There are two different scenarios to evaluate:
1. Mary extracts a white ball, and John extracts a blue ball. The probability of this occurring is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).
2. Mary extracts a blue ball, and John extracts another blue ball. The probability of this happening is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).
The overall probability that John will extract a blue ball is \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56}=\frac{5}{8}\).
Alternative Explanation
The initial probability of drawing a blue ball is \(\frac{5}{8}\). Without knowing the outcomes of previous drawings, the probability of drawing a blue ball stays the same for any successive drawing: second, third, fourth, and so on. There's simply no reason to assume that one drawing is different from another when we're unaware of the outcomes of previous drawings.
If Mary were to extract not just 1 but 7 balls randomly and retain them, even then, the probability that the last ball left for John is blue would still be \(\frac{5}{8}\).
Answer: D
In a basket, there are 3 white balls and 5 blue balls. If Mary draws one ball at random and keeps it, what is the probability that John will subsequently draw a blue ball?
A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)
There are two different scenarios to evaluate:
1. Mary extracts a white ball, and John extracts a blue ball. The probability of this occurring is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).
2. Mary extracts a blue ball, and John extracts another blue ball. The probability of this happening is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).
The overall probability that John will extract a blue ball is \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56}=\frac{5}{8}\).
Alternative Explanation
The initial probability of drawing a blue ball is \(\frac{5}{8}\). Without knowing the outcomes of previous drawings, the probability of drawing a blue ball stays the same for any successive drawing: second, third, fourth, and so on. There's simply no reason to assume that one drawing is different from another when we're unaware of the outcomes of previous drawings.
If Mary were to extract not just 1 but 7 balls randomly and retain them, even then, the probability that the last ball left for John is blue would still be \(\frac{5}{8}\).
Answer: D
General Discussion
04 Jul 2016, 19:59
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I'm confused. Can someone help me here?
Probability = Favourable Outcomes/Total possible outcomes
In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
Probability = Favourable Outcomes/Total possible outcomes
In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
