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25 Mar 2025, 00:30
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Difficulty:
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65% (01:32) correct
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A jar contains 5 blue marbles, 4 red marbles and 3 white marbles. If three marbles are chosen from the jar at random without replacement, what is the probability that each one of the colors will be chosen?
A. 5/144
B. 1/22
C. 3/22
D. 3/11
E. 1/3
A. 5/144
B. 1/22
C. 3/22
D. 3/11
E. 1/3
25 Mar 2025, 00:53
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Total marbles in the jar: \(5 + 4 + 3 = 12\)
We need to find the probability of selecting **one blue, one red, and one white** in any order.
Select 1 blue from 5
\(\frac{5}{12}\)
Select 1 red from 4 (now 11 marbles left)
\(\frac{4}{11}\)
Select 1 white from 3 (now 10 marbles left)
\(\frac{3}{10}\)
Probability for this specific sequence (Blue → Red → White):
\(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\)
Since the selection can occur in any order, there are \(3! = 6\) ways to arrange the three colors.
Final probability:
\(6 \times \frac{1}{22} = \frac{6}{22} = \frac{3}{11}\)
Final Answer: \(\frac{3}{11}\)
We need to find the probability of selecting **one blue, one red, and one white** in any order.
Select 1 blue from 5
\(\frac{5}{12}\)
Select 1 red from 4 (now 11 marbles left)
\(\frac{4}{11}\)
Select 1 white from 3 (now 10 marbles left)
\(\frac{3}{10}\)
Probability for this specific sequence (Blue → Red → White):
\(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\)
Since the selection can occur in any order, there are \(3! = 6\) ways to arrange the three colors.
Final probability:
\(6 \times \frac{1}{22} = \frac{6}{22} = \frac{3}{11}\)
Final Answer: \(\frac{3}{11}\)
25 Mar 2025, 01:09
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We are looking for the probability that each one of colors is chosen without replacement.
We have 5 blue marbles, 4 red marbles, and 3 white marbles.
Choosing a blue, red, and white marble without replacement:
Since the probability of choosing any color combination is equal, it does not matter what scenario we begin with.
P(BRW) = 5/12 * 4/11 * 3/10 = 1/22
We must remember that BRW is just one way to select the color order. There will be a total 3! ways to select all possible color orders, so we multiple the probability of one color combination by 3!
1/22 * 3! = 1/22 * 6 = 3/11
Alternative solution:
We can use combinations to solve the probability as well.
P(each color) = # of favorable events / #number of total events
P(each color) = 5C1 * 4C1 * 3C1 / 12C3
5 * 4 * 3 / 12 * 11 * 10 / 3 * 2 * 1 =
5*4*3*6 / 12*11*10
= 3/11
Option D
We have 5 blue marbles, 4 red marbles, and 3 white marbles.
Choosing a blue, red, and white marble without replacement:
Since the probability of choosing any color combination is equal, it does not matter what scenario we begin with.
P(BRW) = 5/12 * 4/11 * 3/10 = 1/22
We must remember that BRW is just one way to select the color order. There will be a total 3! ways to select all possible color orders, so we multiple the probability of one color combination by 3!
1/22 * 3! = 1/22 * 6 = 3/11
Alternative solution:
We can use combinations to solve the probability as well.
P(each color) = # of favorable events / #number of total events
P(each color) = 5C1 * 4C1 * 3C1 / 12C3
5 * 4 * 3 / 12 * 11 * 10 / 3 * 2 * 1 =
5*4*3*6 / 12*11*10
= 3/11
Option D
