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 March 22nd, 2012, 01:47 PM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Probability, marbles Hi, can someone check this and tell me if I am doing this correctly? I would really appreciate it! Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen. a) Find the probability distribution of X. b) Find the mean of the random variable X ok so for part a i did: P(0B) (6 choose 0) 1*(0.10)^0 (0.3)^6 = 0.000729 P(1B) (6 choose 1) 6* (0.10)^1 (0.3)^5 = .001458 P(2B) (6 choose 2) 15 *(0.10)^2 (0.3)^4 = .001215 P(3B) (6 choose 3) 20 *(0.10)^3 (0.3)^3 = 0.00054 P(4B) (6 choose 4) 15 * (0.10)^4 (0.3)^2 = 0.000135 P(5B) (6 choose 5) 6 * (0.10)^5 (0.3)^1 = 0.00000180 P(6B) (6 choose 6) 1 * (0.10)^6 (0.3)^0 = 0.00000010 Is this right? Or am I doing this completely wrong? Some of the numbers I calculated aren't coming out right.
 March 23rd, 2012, 10:43 AM #2 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Re: Probability, marbles I tried it again and I still am getting wrong answers
March 23rd, 2012, 07:42 PM   #3
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Re: Probability, marbles

Quote:
 Originally Posted by Ataloss Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen. a) Find the probability distribution of X. b) Find the mean of the random variable X
Are you absolutely sure that the marbles are chosen without replacement?
If that's the case, I see that the event of drawing blue marbles are dependent, and the probability of success (i.e. to get blue marbles) is not the same for each draw. In other words, it can't constitute any probability distribution, don't you think???

March 24th, 2012, 12:55 PM   #4
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Re: Probability, marbles

Hello, Ataloss!

Quote:
 Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen. a) Find the probability distribution of X.

There are 8 Blue marbles and 13 Others.

$\begin{array}{cccccccc}
\text{0 Blue (6 Others):} & {6\choose0}\,\left(\frac{13}{21}\cdot\frac{12}{20} \cdot\frac{11}{19}\cdot\frac{10}{18}\cdot\frac{9}{ 17}\cdot\frac{8}{16}\right) &=& \frac{143}{4522} & =& 0.031623176 \\ \\ \\

\text{1 Blue (5 Others):} & {6\choose1}\,\left(\frac{8}{21}\right)\cdot\left(\ frac{13}{20}\cdot\frac{12}{19}\cdot\frac{11}{18}\c dot\frac{10}{17}\cdot\frac{9}{16}\right) &=& \frac{858}{4522} &=& 0.189739054 \\ \\ \\

\text{2 Blue (4 Others):} & {6\choose2}\,\left(\frac{8}{21}\cdot\frac{7}{20}\r ight)\cdot\left(\frac{13}{19}\cdot\frac{12}{18}\cd ot\frac{11}{17}\cdot\frac{10}{16}\right) &=& \frac{2145}{5814} &=& 0.368937049 \\ \\ \\

\text{3 Blue (3 Others):} &{6\choose3}\,\left(\frac{8}{21} \cdot \frac{7}{20} \cdot\frac{6}{19}\right)\cdot\left(\frac{13}{18}\c dot\frac{12}{17}\cdot\frac{11}{16}\right) &=& \frac{2860}{9690} &=& 0.295149639 \\ \\ \\

\text{4 Blue (2 Others):} & {6\choose4}\,\left(\frac{8}{21}\cdot\frac{7}{20}\c dot\frac{6}{19}\cdot\frac{5}{18}\right)\cdot\left( \frac{13}{17}\cdot\frac{12}{16}\right) &=& \frac{195}{1938} &=& 0.100619195 \\ \\ \\

\text{5 Blue (1 Other):} & {6\choose5}\,\left(\frac{8}{21}\cdot\frac{7}{20}\c dot\frac{6}{19}\cdot\frac{5}{18}\cdot\frac{4}{17}\ right)\cdot\left(\frac{13}{16}\right) &=& \frac{78}{5814} &=& 0.013415893 \\ \\ \\

\text{6 Blue (0 Others):} & {6\choose6}\,\left(\frac{8}{21}\cdot\frac{7}{20}\c dot\frac{6}{19}\cdot\frac{6}{18}\cdot\frac{4}{17}\ cdot\frac{3}{16}\right) &=& \frac{1}{1938} &=& 0.000515996 \end{array}$

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