SOLUTION: An urn contains three red and five white balls. a ball is drawn at random, its color is noted,and its replaced along with another ball of the same color. this process is repeated

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Question 1117689: An urn contains three red and five white balls. a ball is drawn at random,
its color is noted,and its replaced along with another ball of the same
color. this process is repeated until three balls have been drawn. find the
mean and standard deviation of the number of red balls.

Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!

There are 8 possible ways to procede:

RRR->(3R,5W)->draw R->then (4R,5W)->then draw R->then (5R,5W)->then draw R
RRW->(3R,5W)->draw R->then (4R,5W)->then draw R->then (5R,5W)->then draw W
RWR->(3R,5W)->draw R->then (4R,5W)->then draw W->then (4R,6W)->then draw R
RWW->(3R,5W)->draw R->then (4R,5W)->then draw W->then (4R,6W)->then draw W
WRR->(3R,5W)->draw W->then (3R,6W)->then draw R->then (4R,6W)->then draw R
WRW->(3R,5W)->draw W->then (3R,6W)->then draw R->then (4R,6W)->then draw W
WWR->(3R,5W)->draw W->then (3R,6W)->then draw W->then (3R,7W)->then draw R
WWW->(3R,5W)->draw W->then (3R,6W)->then draw W->then (3R,7W)->then draw W

Here are the probabilities of each of those procedures:

P(RRR) = (3/8)(4/9)(5/10) =  60/720 = 1/12 <-- 3 red 
P(RRW) = (3/8)(4/9)(5/10) =  60/720 = 1/12 <-- 2 red
P(RWR) = (3/8)(5/9)(4/10) =  60/720 = 1/12 <-- 2 red
P(RWW) = (3/8)(5/9)(6/10) =  90/720 = 1/8  <-- 1 red
P(WRR) = (5/8)(3/9)(4/10) =  60/720 = 1/12 <-- 2 red
P(WRW) = (5/8)(3/9)(6/10) =  90/720 = 1/8  <-- 1 red
P(WWR) = (5/8)(6/9)(3/10) =  90/720 = 1/8  <-- 1 red
P(WWW) = (5/8)(6/9)(7/10) = 210/720 = 7/24 <-- 0 red
---------------------------------------------------
                      sum = 720/720 = 1
x   P(x red)
------------------------------------------ 
0   P(0 red) = 7/24
1   P(1 red) = 1/8+1/8+1/8 = 3/8
2   P(2 red) = 1/12+1/12+1/12 = 3/12 = 1/4
3   P(3 red) = 1/12

[Observe as a check that those probabilities have sum 1 
 7/24+3/8+1/4+1/12 = 7/24+9/24+6/24+2/24 = 24/24 = 1]

mean = µ = E(X) = Σx∙p(x) = (0)(7/24) + (1)(3/8) + 2(1/4) + 3(1/12) = 
                                 0    +   3/8    +   1/2  +    1/4  =
                                          3/8    +   4/8  +    2/8  =
                                                     9/8

variance = σ² = E(X²) - [E(X)]² 

E(X²) =  Σx²∙p(x) = (0)²(7/24) + (1)²(3/8) + 2²(1/4) + 3²(1/12) =
                         0     +   3/8     +  4(1/4) +  9(1/12) =
                                   3/8     +     1   +    3/4   =
                                   3/8     +    8/8  +    6/8   =
                                               17/8   

variance =  σ² = E(X²) - [E(X)]² = 17/8 - [9/8]² = 81/64 =  

                                   17/8 - 81/64

                                 136/64 - 81/64

                                     55/64    

standard deviation = σ = √(σ²) = √[55/64]  √(55)/8  = 0.9270 

Edwin