Author Topic: probability distribution (Read 625 times) Tweet Share

scatteam_2 · « **on:** September 30, 2010, 07:37:35 am »

3 balls are selected from a jar containing 4 black balls and 6 red balls. Find the probability distribution of the number of black balls in the sample if:

a) the ball chosen is replaced after each selection
b) the ball chosen is not replaced after each selection

step by step explanation? im stuck

the.watchman · « **Reply #1 on:** September 30, 2010, 07:54:43 am »

REPLACEMENT:

$Pr(B=0) = Pr(R=3) = (\frac{6}{10})^3 = \frac{27}{125}$

$Pr(B=1) = Pr(R=2) = 3 \cdot (\frac{6}{10})^2 \cdot \frac{4}{10} = \frac{18}{125}$

The three times above indicates that there are three different ways of choosing one black (BRR, RBR, RRB)
Continue in this fashion, where 6/10 is the Pr of getting a red and 4/10 is the Pr of getting a black

NO REPLACEMENT:

You need to consider that when a ball is taken out: (a) the number of balls of that colour decreases and (b) the total number of balls decreases as well

$Pr(B=0) = Pr(R=3) = \frac{6}{10} \cdot \frac{5}{9} \cdot \frac{4}{8} = \frac{1}{6}$

$Pr(B=1) = Pr(R=2) = 3 \cdot \frac{4}{10} \cdot \frac{6}{9} \cdot \frac{5}{8} = \frac{1}{2}$

For B=1 and B=2, you might think that the order would mean that BRR, RBR, and RRB are different in terms of their probabilities but you'll find that that is not true. You can still find the Pr of one scenario and multiply it by the numbers of ways.

Hope this helps

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Author Topic: probability distribution (Read 625 times) Tweet Share

scatteam_2

probability distribution

the.watchman

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