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.

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

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

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

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).

Your solution is almost correct. Here's the revised solution:

a) To create a probability distribution table, you need to find the probability of choosing each possible number of red jelly beans.

- Prob(0 red) = Number of ways to choose 0 red beans / Total number of ways to choose 4 beans
= C(7, 4) / C(15, 4)
= 35 / 1365
≈ 0.0256

- Prob(1 red) = Number of ways to choose 1 red bean * Number of ways to choose 3 green beans / Total number of ways to choose 4 beans
= (C(8, 1) * C(7, 3)) / C(15, 4)
= (8 * 35) / 1365
≈ 0.2068

- Prob(2 red) = Number of ways to choose 2 red beans * Number of ways to choose 2 green beans / Total number of ways to choose 4 beans
= (C(8, 2) * C(7, 2)) / C(15, 4)
= (28 * 21) / 1365
≈ 0.3444

- Prob(3 red) = Number of ways to choose 3 red beans * Number of ways to choose 1 green bean / Total number of ways to choose 4 beans
= (C(8, 3) * C(7, 1)) / C(15, 4)
= (56 * 7) / 1365
≈ 0.2378

- Prob(4 red) = Number of ways to choose 4 red beans / Total number of ways to choose 4 beans
= C(8, 4) / C(15, 4)
= 70 / 1365
≈ 0.0513

So, the probability distribution table for choosing red jelly beans is:
| Number of Red Jelly Beans | Probability |
|--------------------------|-------------|
| 0 | 0.0256 |
| 1 | 0.2068 |
| 2 | 0.3444 |
| 3 | 0.2378 |
| 4 | 0.0513 |

b) The probability that at least 2 chosen jelly beans are green given that they are not all red can be calculated as the sum of the probabilities of choosing 2, 3, or 4 green jelly beans (excluding the case of all 4 red jelly beans).

Probability = Prob(2 green) + Prob(3 green) + Prob(4 green)
= (C(7, 2) * C(8, 2)) / C(15, 4) + (C(7, 3) * C(8, 1)) / C(15, 4) + (C(7, 4) * C(8, 0)) / C(15, 4)
= (21 * 28) / 1365 + (35 * 8) / 1365 + (35 * 1) / 1365
≈ 0.4806

So, the probability that at least 2 chosen jelly beans are green given that they are not all red is approximately 0.4806.

Well done on your solution, but remember to double-check your calculations.