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Consider a lot consisting of 3 blue balls and 1 red ball.

Suppose I pick 2 balls one after another without replacement, now the probability of 2 balls being blue:

$$\frac{3\choose 2}{4 \choose 2}$$

Now taking different approach using conditional probability, the solution is also:

$$\frac34\times\frac23$$

Now it turns out that both of them evaluate to $1/2$.

Why this happens and does it happen always?

Is solving the problem using combinations an efficient approach always?

Or am I missing something basic out here!

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  • $\begingroup$ Without replacement is the keyword here. The combinations approach results in an arrangement without replacement. Hence the answer matches. Try to solve the same problem with replacement=True. $\endgroup$ – Sriharsha Madala Sep 23 '15 at 18:28
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We are talking of drawing without replacement.

For the particular case you have given, the two approaches match, but beware

If the question were probability of getting 1 red, 1 blue

$(a) \dfrac{{3\choose1}{1\choose1}}{4\choose2} = \dfrac12$ , but

$(b) \dfrac34\cdot\dfrac13 = \dfrac14$ and you need to multiply by $2!$ to take into account different orders in which the events can occur.

On the other hand, if the problem asked the probability of a red followed by a blue,

$(b) \dfrac34\cdot\dfrac13 = \dfrac14$ is correct,

and for $(a)$, you need to divide $\dfrac{\binom{3}{1}\binom{1}{1}}{\binom{4}{2}}= \dfrac12$ by $2!$ to get the specified order

Adopting the KISS approach, I use combinations when order is unspecified, and direct multiplication of probabilities when order is specified.

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Both approaches are correct. Both methods can be viewed as combinatorial.

In your first approach, your sample space consists of ${4 \choose 2} = 6$ unordered samples, each with probability $1/6.$ Three of the six outcomes correspond to choosing two blue balls, for a probability of $1/2.$ It is OK to use unordered samples because we are interested in the number of balls chosen of each color, not the order in which choices were made.

Maybe the balls are numbered 1, 2, 3, 4, with the first three being blue. Then outcomes that satisfy your event are $\{12, 13, 23\}$. (I use $numerical$ order in this listing because the $actual$ order of listing is irrelevant.

In your second approach, you can consider that you have a sample space consisting of $12$ ordered outcomes, each with probability $1/12.$ Six of the twelve ordered outcomes have two blue balls, so the probability is $1/2.$

This second approach would be necessary if you wanted to answer a question such as "What is the probability I choose a blue ball first and then the red one?"

Numbering the balls as before, the six outcomes out of twelve that satisfy the event $\{\text{Two Blue}\}$ are $\{12, 13, 23, 21, 31, 32\}.$ Here I am not free to use numerical order in making my list because the actual order of selection does matter.

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This is one more method to solve this problem:
Consider a Lot consists of {B1,B2,B3,R} B:Blue ball R:Red ball.

Now list out all the combinations we get when we choose 2 balls from the Lot, one after another(without replacement).Let the event be S.

S={(B1,B2),(B1,B3),(B1,R),(B2,B1),(B2,B3),(B2,R),(B3,B2),(B3,B1),(B3,R),(R,B1),(R,B2),(R,B3)}.

Now our event E is getting 2 Blue balls,

E={(B1,B2),(B1,B3),(B2,B1),(B2,B3),(B3,B2),(B3,B1)}.

Now probability of getting event E is,
P(E)=n(S)/n(E);which is 6/12=1/2. n:cardinality of a set(ie.number of elements).

Inference: There is hell a lot of way to get solution to a probability question,all the answers match provided the approach is right.Everyone chooses their own approach,so there is no "efficient" approach. This is also a reason why I(Most of us) love it.

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