Stats Help Please!!!

Question

Consider two boxes, first one (B1) contains 6 green (G) and 5 red (R) balls, the second box (B2) contains 4 green (G) and 7 red (R) balls. We toss a fair coin and select one of the boxes from which one ball is taken without replacement, Repeat the coin flip and select one of the boxes and draw the second ball from the box.

a) What is the probability having two red balls if the box selected in the first step was box 1 (B1)?

b) What is the probability of having a red and a green ball at the end (disregard the part a)?

Answer 1

This exercise employs, among other things, the ideas of "mutually exclusive" outcomes and "independent" outcomes. There are also two Big Questions: is there replacement? and, is there order (i.e., does order matter)?

In this case there is NO replacement -- so the probability of choosing "Red" may change from pick to pick.

But there IS "order" -- a draw from (1, 2) is not necessarily considered identical to (2, 1).

There is an element of random "choice," so I can think of 4 possible "choice" scenarios:

box choices: ( Box 1, Box1) ; (1, 2) ; (2, 1) ; or (2, 2)

** NOTE ** A "tree" or branch diagram may come in handy for you -- actually, I would set up TWO branch diagrams, and attach them together at the end -- there are 2 paths that depend on the first choice (flip #1) -- Box 1 or Box 2 -- and 2 outcomes (Red / not Red) per choice: 2x2 = 4, We repeat the process for the 2nd choice, so there are 2x2x2x2 = 16 paths and all of them may lead to different probability values!

Before we choose, let's note that the P(Red) for Box 1 is 5/11, and P(Red) for Box 2 is 7/11.

Choice #1: Flip: H = Box 1, and T = Box 2. (This is arbitrary on my part)

(PART A) We flip "H" and choose from Box 1: P(Red) = 5/11 and P(not Red) = 6/11.

After that, we flip the coin again -- and we choose again from Box 1 if we get "H."

Flip 2: Has FOUR routes, GIVEN THAT we know we chose from Box 1 on the FIRST draw, and also that we are interested in drawing TWO RED balls:

:

Choose from Box 1: P(Red) = 4/10 if we chose Red (path 1) , but 5/10 IF we got Green on the FIRST draw (path 2) ;

Choose from Box 2: P(Red) = 7/11 (path 3) , and 7/11 IF we got Green (path 4).

Now we can lay out the probabilities:

... 2nd draw from Box 1:

Path 1 P(Red) = 5/11 * 4/10 , if we draw Red, we have (R,R) ** good!

Path 2 P(Red) = 5/11 * 6/10 , if we draw Green, we have (R,G)

... 2nd draw from Box 2:

Path 3 P(Red) = 5/11 * 7/11 , if we draw Red, we have (R,R) ** good!

Path 4 P(Red) = 5/11 * 4/11 , if we draw Green, we have (R,G)

... on the remaining paths (#s 5-8), the first choice out of Box 1 would be a GREEN ball...

Outcomes are *mutually exclusive* (AKA "disjoint"); we can simply add the probabilities of the desired outcomes:

Path 1 P(R,R) = 5/11*4/10 =20/110

Path 3 P(R,R) = 5/11*7/11 = 35/121

and finally

(20/110) + (35/121) = (I assume you use a calculator here) = some decimal amount., about 0.18 + 0.29 or 0.47.

I hope this helps, and not only on this exercise. If you have any comments or questions, post them here!

Regards,

:PC

Answer 2

a) P(2R) = P(R from B1) ( 1/2 P(R from B1 again) + 1/2 P(R from B2))

In the notation that follows: P(RB1)1/2(PRB1'+P(RB2))

b) P(R,G) = 1/4 (P(RB1)P(GB1')+P(GB1)P(RB1')) + 2 pulls from B1

1/4(P(RB2)P(GB2')+P(GB2)P(RB2')) + 2 pulls from B2

1/2(P(RB1)P(GB2) + P(RB2)P(GB1) It is twice as likely to get coin toss B1, B2

P(RB1) = 5/11, P(GB1) = 6/11 and P(GB1', red out) = 6/10 P(GB1', red out) = 4/10, etc.

Good luck