A red box has 4 balls and 6 black balls.
A black box has 6 red balls and 4 red balls.
Step 1: Toss a coin. If heads: pick a ball from the red box. If tails: pick a ball from the black box.
Step 2: Choose a ball from the box having the same colour as the ball selected in step 1.

Let X = number of red balls you draw in the 1st two steps. Find E(X).

Question

A red box has 4 balls and 6 black balls.
A black box has 6 red balls and 4 red balls. 
Step 1: Toss a coin. If heads: pick a ball from the red box. If tails: pick a ball from the black box. 
Step 2: Choose a ball from the box having the same colour as the ball selected in step 1.

Let X = number of red balls you draw in the 1st two steps. Find E(X).

kirbykirby · Answer

The solution used indicator variables (sorry the dashes in the cases, i don't know how to add space in LaTeX).R = red, B= black
$
 I_1 = \begin{cases}
        1  & ball-in-step1-is-red\
        0 & otherwise
        \end{cases}
$ 

$
 I_2 = \begin{cases}
        1  & ball-in-step2-is-red\
        0 & otherwise
        \end{cases}
$ 

$X=I_1+I_2$
$E(X)=E(I_1)+E(I_2)$
$E(X)=P(I_1=1)+P(I_2=1)$

They get $P(I_1=1)=0.5$ which I understood, but for $P(I_2=1)$:

$P(I_2=1)=P(2^{nd} ball=R)$
$P(I_2=1)=P(2^{nd}=R|1^{st}=R)*P(1^{st}=R)+P(2^{nd}=R|1^{st}=B)*P(1^{st}=B)$
                $=(4/10)*(1/2)+(6/10)*(1/2)$

I'm having so much trouble how they get 4/10 for P(2nd=R|1st=R)... if you picked a red ball initially from the red box won't the probability be less than 4/10?

kirbykirby · Answer

The probalem doesn't state anything though if balls are being picked with or without replacement

anonymous · Answer

then you can't do it

kirbykirby · Answer

are they doing it correctly if there's assuming with replacement? which is what I think they're doing

kirbykirby · Answer

they're assuming*