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There are x red marbles, y blue marbles, and z yellow marble
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30 Jan 2020, 14:49
30 Jan 2020, 14:49
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There are x red marbles, y blue marbles, and z yellow marbles in a sack. Three marbles are chosen at random, without replacements. If x, y, and z are all at least equal to 3, then which of the following must be true?
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
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Re: There are x red marbles, y blue marbles, and z yellow marble
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01 Feb 2020, 08:17
01 Feb 2020, 08:17
2
Carcass wrote:
There are x red marbles, y blue marbles, and z yellow marbles in a sack. Three marbles are chosen at random, without replacements. If x, y, and z are all at least equal to 3, then which of the following must be true?
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
There are x red marbles, y blue marbles, and z yellow marbles in a sack
thus, total marbles = \((x + y + z)\)
Let us calculate each of the options:
A. The probability of drawing exactly 3 red marbles
= Probability that the 1st marble is red AND the 2nd marble is red AND the 3rd marble is red
Probability of picking the 1st red marble = (# red marbles)/(# total marbles) = \(\frac{x}{(x + y + z)}\)
After picking one red marble, the number of red marbles would be \((x - 1)\) and total marbles = \((x + y + z - 1)\)
Thus, probability of picking the 2nd red marble = (# red marbles)/(# total marbles) = \(\frac{(x - 1)}{(x + y + z - 1)}\)
After picking one more red marble, the number of red marbles would be \((x - 2)\) and total marbles = \((x + y + z - 2)\)
Thus, probability of picking the 3rd red marble = (# red marbles)/(# total marbles) = \(\frac{(x - 2)}{(x + y + z - 2)}\)
Thus, probability of drawing exactly 3 red marbles
= \(\frac{x}{(x + y + z)}\) * \(\frac{(x - 1)}{(x + y + z - 1)}\) * \(\frac{(x - 2)}{(x + y + z - 2)}\)
Thus, A is correct
B. The probability of drawing at least 1 red marble
= 1 - Probability that no marble is red
= 1 - (Probability that the 1st marble is NOT red AND the 2nd marble is NOT red AND the 3rd marble is NOT red)
= 1 - [(Probability that 1st marble is blue or yellow) * (Probability that 2nd marble is blue or yellow) * (Probability that 3rd marble is blue or yellow)]
= \(1 - (\frac{(y+z)}{(x + y + z)}) * (\frac{(y+z - 1)}{(x + y + z - 1)}) * (\frac{(y+z - 2)}{(x + y + z - 2)})\)
Thus, B is correct
C. In the same way as option A: The probability of drawing exactly 3 blue marbles
= Probability that the 1st marble is blue AND the 2nd marble is blue AND the 3rd marble is blue
= \(\frac{y}{(x + y + z)}\) * \(\frac{(y - 1)}{(x + y + z - 1)}\) * \(\frac{(y - 2)}{(x + y + z - 2)}\)
Thus, C is not correct
Answer: A,B
Re: There are x red marbles, y blue marbles, and z yellow marble
[#permalink]
11 Aug 2020, 09:45
11 Aug 2020, 09:45
2
Carcass wrote:
There are x red marbles, y blue marbles, and z yellow marbles in a sack. Three marbles are chosen at random, without replacements. If x, y, and z are all at least equal to 3, then which of the following must be true?
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
Indicate all possible values.
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
\(\square\) The probability of drawing at least 1 red marble is
\(1- \left[ \begin{array}{cc|r} \frac{y+z}{x+y+z} \times \frac{y+z-1}{x+y+z-1} \times \frac{y+z-2}{x+y+z-2} \end{array} \right]\)
\(\square\) The probability of drawing exactly 3 blue marbles is
\(\frac{y}{x+y+z} \times \frac{y-x}{x+y+z-1} \times \frac{y+z}{x+y+z-2}\)
\(\square\) The probability of drawing exactly 3 red marbles is
\(\frac{x}{x+y+x} \times \frac{x-1}{x+y+z-1} \times \frac{x-2}{x+y+z-2}\)
Since \(x,y,z >= 3\), let \(x\) = 10 , \(y\) = 3, and \(z\) = 3
Then we get:
\(\frac{10}{10+3+10}\) \(\times\) \(\frac{10-1}{10+3+3-1}\) \(\times\) \(\frac{10-2}{10+3+3-2}\)
This gives:
\(\frac{10}{23}\) \(\times\) \(\frac{9}{15}\) \(\times\) \(\frac{8}{14}\)
When it should be:
\(\frac{10}{16}\) \(\times\) \(\frac{9}{15}\) \(\times\) \(\frac{8}{14}\)
So A doesn't work? Have I thought about this incorrectly?
