In a bag there are 8 red jelly beans and 7 green jelly beans. The 4 jelly beans are randomly chosen from the bag without replacement.

Question

a)create a probability distribution table for choosing red jelly beans
b)what is the probability that at least 2 chosen jelly beans are green given that they are not all red?
Thanks.
Please check if my solution is correct:
a) prob(1)=8/15
prob(2)=7/14
prob(3)=4/15
prob(all 4 red)=8/525
b)7/15x6/14=1/5

Answer 1

For (a), you may have double counted some cases, since the total is 1381/1050 >1, whichis impossible.

Answer 2

I'm really confused now. Can you please explain where I've double counted? Thanks in advance.

Answer 3

Actually, the probability table should include the case P(0) as well.

Do you know how to make a tree diagram? This is a foolproof way to make a distribution table, as long as it is not too complicated.
In this case, you will have 2^4=16 "leaves", i.e. where the branches end.

If you proceed by counting, the following is a possibility:

P(0)=P(GGGG)=(7/15)(6/14)(5/13)(4/12)
=(7*6*5*4)/(15*14*13*12)
=840/32760
Do not simplify so we can add up later on. The denominator is always 32760, so I will work on the numerator only.

P(1)=
[arrangements:
RGGG 8*7*6*5
GRGG 7*8*6*5
GGRG 7*6*8*5
GGGR 7*6*5*8]
=4(8*7*6*5)
=6720/32760

P(2)=
[arrangements]
RRGG 8*7*7*6
RGRG
RGGR
GRRG
GRGR
GGRR]
=6*(8*7*7*6)
=14112/32760

P(3)=
[arrangements
RRRG 8*7*6*7
RRGR
RGRR
GRRR]
=9408/32760

P(4)=
[arrangements
RRRR]
=8*7*6*5/32760
=1*1680/32760
=1680/32760

So the probability distribution table for the number of red jelly beans when 4 are picked:
0 840/32760
1 6720/32760
2 14112/32760
3 9408/32760
4 1680/32760

Check: add up all numerators to see that they add up to 32760:
840+6720+14112+9408+1680=32760 √

Part (B) can be inferred from the results of part (A).