discrete prob dist

[Sonal7]

I have a question about part c. I am not sure how to work this one out as the dist will be infinite. The Q says the procedure is repeated until a red is removed. There are no other conditions so this could go on for ever.

 

[nasi112]

Let me try to explain the situation

First, remember this it was said colour is noted and token is replaced means probability with REPLACEMENT

[MATH]Y[/MATH] is the number of green tokens

[MATH]Y = 0[/MATH]Red
means we removed a red token from the first time
[MATH]p(Y=0) = \frac{3}{10}[/MATH]
[MATH]Y = 1[/MATH]Green, then Red
[MATH]p(Y = 1) = \frac{7}{10} \cdot \frac{3}{10} = \frac{21}{100}[/MATH]
[MATH]Y = 2[/MATH]Green, Green, then Red
[MATH]p(Y = 2) = \frac{7}{10} \cdot \frac{7}{10} \cdot \frac{3}{10} = \frac{147}{1000}[/MATH]
[MATH]Y = 3[/MATH]Green, Green, Green, then Red
[MATH]p(Y = 3) = \frac{7}{10} \cdot \frac{7}{10} \cdot \frac{7}{10} \cdot \frac{3}{10} = \frac{1029}{10000}[/MATH]
repeat this process until [MATH]Y = 7[/MATH]
Note that, when [MATH]Y = 7[/MATH], you don't need to remove a Red token because you already have all Green tokens (Maximum).

Once you got all the probabilities, you can find [MATH]VAR(Y)[/MATH] easily.

 

[Sonal7]

thank you - I need to think about a bit more about the technical terms used.