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09 Aug 2024, 01:30
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D
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Two chocolates need to be picked from a box with replacement. If all the chocolates in the box are wrapped in either green or blue wrapper. Then what's the probability that we get one chocolate of each color.
(1) The probability of getting chocolates of the same color is 5/9
(2) The number of chocolates with a blue wrapper is twice the number of chocolates with a green wrapper
(1) The probability of getting chocolates of the same color is 5/9
(2) The number of chocolates with a blue wrapper is twice the number of chocolates with a green wrapper
10 Aug 2024, 11:55
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Bunuel
(1) The probability of getting chocolates of the same color is 5/9
Prob. of not getting same color = \(1- \frac{5}{9} = \frac{4}{9}\)
SUFF.
(2) The number of chocolates with a blue wrapper is twice the number of chocolates with a green wrapper
Let it be \(x\) and \(2x\) respectively.
Therefore \(\frac{x}{3x}*\frac{2x}{3x}*2! =\frac{4}{9}\) We multiply by \(2\) because we don't know which color will be picked first. We could pick BG or GB.
SUFF.
Ans D
Hope it helped.
17 Aug 2024, 03:40
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Two chocolates need to be picked from a box with replacement
(1) The probability of getting chocolates of the same color is \(\frac{5}{9}\)
=> The probability that we get one chocolate of each color = \(1 - \frac{5}{9} = \frac{4}{9}\)
(2) The number of chocolates with a blue wrapper is twice the number of chocolates with a green wrapper
B = 2G => Total (B+G) = 3G
\(\frac{B}{B+G} = \frac{2}{3}\)
\(\frac{G}{B+G} = \frac{1}{3}\)
The probability of getting chocolates of the same color
= \((\frac{B}{B+G})^2 + (\frac{G}{B+G})^2\)
= \((\frac{2}{3})^2 + (\frac{1}{3})^2\)
= \(\frac{5}{9}\)
Like statement 1, we can infer the probability that we get one chocolate of each color
(1) The probability of getting chocolates of the same color is \(\frac{5}{9}\)
=> The probability that we get one chocolate of each color = \(1 - \frac{5}{9} = \frac{4}{9}\)
(2) The number of chocolates with a blue wrapper is twice the number of chocolates with a green wrapper
B = 2G => Total (B+G) = 3G
\(\frac{B}{B+G} = \frac{2}{3}\)
\(\frac{G}{B+G} = \frac{1}{3}\)
The probability of getting chocolates of the same color
= \((\frac{B}{B+G})^2 + (\frac{G}{B+G})^2\)
= \((\frac{2}{3})^2 + (\frac{1}{3})^2\)
= \(\frac{5}{9}\)
Like statement 1, we can infer the probability that we get one chocolate of each color
