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29 Apr 2024, 01:30
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C
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:
65% (01:14) correct
35%
(01:16)
wrong
based on 131
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A bag of marbles consist of a mixture of black and red marbles. What is the probability of choosing a red marble followed by a black marble without a replacement ?
(1) The probability of choosing a black marble first is 1/3.
(2) There are 10 black marbles in the bag.
(1) The probability of choosing a black marble first is 1/3.
(2) There are 10 black marbles in the bag.
29 Apr 2024, 06:25
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Let the number of black marbles = \(b\) and the number of red marbles = \(r\).
The probability of choosing a red marble followed by a black marble will therefore be: \(\frac{r}{b+r}*\frac{b}{b+r-1}\)
(1) The probability of choosing a black marble first is 1/3.
From this one gets that the ratio of black to red marbles is \(1:2\), alternatively that \(r = 2b\). However, without the marbles being replaced with the same colour it is impossible to solve for the probability of choosing a red marble followed by a black marble.
Looking at \(\frac{r}{b+r}*\frac{b}{b+r-1}\), even if one were to plug in \(r = 2b\), one will not be able to solve for a numeric value given the subtraction sign in the denominator in \(\frac{b}{b+r-1}\).
To demonstrate this, plugging in \(r = 2b\): \(\frac{2b}{b+2b}*\frac{b}{b+2b-1}\)
\(\frac{2b}{3b}*\frac{b}{3b-1}\)
\(\frac{2}{3}*\frac{b}{3b-1}\)
\(\frac{2b}{9b-3}\)
INSUFFICIENT
(2) There are 10 black marbles in the bag.
Without knowing how many red marbles are in the bag, it is impossible to solve the question.
INSUFFICIENT
(1+2)
Putting them together, we know that there are 10 black marbles and therefore 20 red marbles which is enough info to solve this question.
\(\frac{20}{30}*\frac{10}{29}\)
\(\frac{20}{87}\)
SUFFICIENT
ANSWER C
The probability of choosing a red marble followed by a black marble will therefore be: \(\frac{r}{b+r}*\frac{b}{b+r-1}\)
(1) The probability of choosing a black marble first is 1/3.
From this one gets that the ratio of black to red marbles is \(1:2\), alternatively that \(r = 2b\). However, without the marbles being replaced with the same colour it is impossible to solve for the probability of choosing a red marble followed by a black marble.
Looking at \(\frac{r}{b+r}*\frac{b}{b+r-1}\), even if one were to plug in \(r = 2b\), one will not be able to solve for a numeric value given the subtraction sign in the denominator in \(\frac{b}{b+r-1}\).
To demonstrate this, plugging in \(r = 2b\): \(\frac{2b}{b+2b}*\frac{b}{b+2b-1}\)
\(\frac{2b}{3b}*\frac{b}{3b-1}\)
\(\frac{2}{3}*\frac{b}{3b-1}\)
\(\frac{2b}{9b-3}\)
INSUFFICIENT
(2) There are 10 black marbles in the bag.
Without knowing how many red marbles are in the bag, it is impossible to solve the question.
INSUFFICIENT
(1+2)
Putting them together, we know that there are 10 black marbles and therefore 20 red marbles which is enough info to solve this question.
\(\frac{20}{30}*\frac{10}{29}\)
\(\frac{20}{87}\)
SUFFICIENT
ANSWER C
19 Jul 2025, 00:32
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