Difference between methods for sampling without replacement

Question

A bag contains 2 red balls, 3 red cubes, 3 blue balls and 2 blue cubes. A person picks 3 objects at random from the bag and does not replace them.

What is the probability that the person picks 2 blue balls?

My initial idea to solve this was using unordered sampling without replacement, although I tend not to think of these problems in this way and instead just use combinations and permutations, but that gave me an answer I feel was wrong:

Our sample space size is $10 \choose 3$ since we are selecting 3 things from a set of 10 things. We want to select 2 blue balls from a possible total of three balls so that gives us $3 \choose 2$ and we then select the third and final object from a set of 8 remaining objects which gives us $8 \choose 1$.

Finally we have $\mathbb{P}$(2 blue balls)$=\frac{{3 \choose 2} {8 \choose 1}}{10 \choose 3}=\frac{1}{5}$.

This value seems too high to me for it to be correct, so I tried the following method:

Sample space size is $\frac{10!}{(10-3)!}=720$ and the event two blue balls is $\frac{3!}{(3-2)!}=6$

Hence $\mathbb{P}$(2 blue balls)$=\frac{6}{720}=\frac{1}{120}$ and this seems more correct to me.

Could someone please tell me which answer is correct, and why both unordered and ordered sampling don't give the same answer?

Also, following on from the first Q, what is the probability that all the objects drawn are blue?

We have 5 blue objects in total.

Probability of picking first object blue is $5 \choose 1$

Probability of picking second object blue is $5 \choose 2$

Probability of picking third object blue is $3 \choose 1$

Then $\mathbb{P}$(all blue)$=\frac{5 \times 4 \times 3}{10 \choose 3}$

Is this correct?

Thanks!!

Answer 1

Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.

In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.

I will leave it to you to show that the probability of "not a blue ball, blue ball, blue ball" in that order is also 1/120. So the probability of two blue balls out of three is 3/120= 1/40.

Answer 2

Your approach to calculating the probability was not wrong, you made a simple mistake. You chose 2 blue balls, from the total of 3 blue balls, i.e 3C2, then you chose 1 final item from 8 remaining items. However, one of those 8 items is also a blue ball, so if you picked it you would have 3 blue balls, not 2. You should pick from the 7 remaining non-blue-ball items. So the probability is

$$ P(\text{2 blue balls}) = \frac{\binom{3}{2}\binom{7}{1}}{\binom{10}{3}} = 7/40 $$

You can also see this in the way user247327 explained. You have 3/10 probability to pick the first blue ball, then 2/9 probability to pick the second, and 7/8 ways to pick the last. $(3/10)(2/9)(7/8) = 7/120$. Then there are three orderings of the ball, so $P = 7/120 \cdot 3 = 7/40$.