A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red?

I disagree with the given answer $\frac29$. Can someone please point why my solution is wrong?

P(Both Red) + P(Second is Red):

P(Both Red) $=\frac{\binom22}{\binom92}$

P(Second is Red) $=\frac{\binom71\binom21}{\binom92}$

This comes out to be $\frac5{12}$.


Total choices = 9
Red balls = 2
P(Both red) = P(1st red) $\times$ P(2nd red) = $\dfrac29 \times \dfrac18$
P(only 2nd ball red) = $\dfrac79 \times \dfrac28$

Required probability = P(Both red) + P(only 2nd ball red) $= \dfrac{16}{9\times8} = \dfrac29$


1) Your argument is wrong since you have over-counted the event the second marble is red.

$P(2nd\, is\, R) = P(both\, are\, R) + P(1st\, is\, not\, R\, and\, 2nd\, is\, R)$

$P(both\, are\, R) = \frac{2}{9}\frac{1}{8} = \frac{2}{72}$

$P(1st\, is\, not\, R\, and\, 2nd\, is\, R) = \frac{7}{9}\frac{2}{8} = \frac{14}{72}$


$P(2nd\, is\, R) = \frac{2}{72} + \frac{14}{72} = \frac{16}{72} = \frac{2}{9}$

2) By symmetry of the problem you can tell immediately the solution is $\frac{2}{9}$.

The problem is the same as you have 9 balls (3W, 4B, 2R). You arrange them randomly in a line. What is the probability of the second ball is R? The probability of the second ball is R is the same as the probability of any particular ball is R (just think of swap the position of the second ball to any other ball - they have equal chance). And the probability of any ball is R is $\frac{2}{9}$ (just think as the first ball chosen they have 2 out of 9 chance to be Red - and then swap position of the first ball to any particular ball).


You have overcounted the cases where only the second marble drawn is red.

The correct answer can be simply derived by noting that the draw here is equivalent to picking the first two objects in a random permutation of the marbles, where any specific position has a $\frac29$ chance of being red.

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    $\begingroup$ I really like it to think in this way. It is very easy to see that the probability of kth ball drawn is red is $\dfrac29$. $\endgroup$ – Rahul Goswami Jul 25 '18 at 7:36
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    $\begingroup$ Actually, the OP over counted the case in which only the second marble is red by not taking the order of selection into account. $\endgroup$ – N. F. Taussig Jul 25 '18 at 8:06

It should be immediately obvious, without any calculation, that the answer is $2/9$.

There are 9 marbles. One of them will be the lucky one that is the second to be picked. The problem is symmetrical - the marbles are identical for the purposes of randomly drawing them, so there's no reason that one of them will be more likely than the others to be the lucky marble.

So they all have the same chance to be picked second (just like they all have the same chance to be picked first, or fifth, or last). There are 9 of them, so each has a probability of $1/9$. Two of them are red, so the probability that the second marble is red is $2/9$.

Let that sink in. The above is the correct way to solve the problem. Jumping mindlessly into needless calculations is often a symptom of poor understanding of the problem.

But if you insist on some unnecessary calculations, you can also solve this by conditioning on the first marble picked.

There are two mutually exclusive scenarios:

  1. The first marble is red (happens with probability $2/9$). Given this, the probability the the second marble will be red is $1/8$ (since there is one red marble left, out of 8).

  2. The first marble is not red (happens with probability $7/9$). Given this, the probability that the second marble will be red is $2/8$.

So the total probability that the second marble is red is $\frac29\cdot\frac18+\frac79\cdot\frac28=\frac2{72}+\frac{14}{72}=\frac{16}{72}=\frac29$.

As for why your solution is wrong - I can only direct the question back at you and ask, why do you think it is right?

  • $\begingroup$ Hi Meni! figures you'd be here :) $\endgroup$ – Vinnie Falco Jul 25 '18 at 14:07
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    $\begingroup$ @VinnieFalco: I'm not really here, it's merely an optical illusion. $\endgroup$ – Meni Rosenfeld Jul 25 '18 at 14:33

To gain intuition on this:

suppose there are $9$ spots labeled with numbers $1,2,\dots,9$.

Randomly on every spot a marble is placed.

What is the probability that on spot $2$ a red marble is placed?

$\frac29$ of course, since there are $9$ equiprobable candidates for spot $2$ and exactly $2$ of them are red.

Drawing without replacement can be interpreted like this.

If you only draw $2$ of the $9$ marbles then you do exactly the same, but only spots $1$ and $2$ are taken into account.


You write P(Both Red) + P(Second is Red)

But this is a bit hard to parse. Since P(Second is Red) is explicitly the probability that the problem asks about, so you probably mean something else.

You go on to say;

P(Second is Red) = 7C1 * 2C1/ 9C2 And the question becomes, what is this calculation supposed to represent.

In the context of the problem 7C1*2C1/9C2 is the probablity that one of the red balls is one first two while the other is one of the last 7, ie. exactly one of the first two balls are red.

We could do the calculation P(Second is Red)=P(Both Red)+P(Exactly one of the first two are red)*P(Second is Red|Exactly one of the first two are red)=2C2/9C2+(7C1*2C1/9C2)*1/2=2/9

A more straigthforward calculation. And one that you where most likely after is the probability that the red balls are in position 2 white the other is in some position after this. This then would be 7C1*1C1/9C2.


What other answers don't point out is why your reasoning is wrong in the first place.

The problem is that $\binom92$ is for an unordered set of 2 elements taken from 9-element set. But in your case you have to have an ordered set since only the second element of the choice is relevant and so the order is relevant.

The correct number of possibilities in such case for m elements chosen from n-element set is $\frac{n!}{(n-m)!}$ (and respectively for your needs:)

  • $\frac{9!}{(9-2)!} = 9*8$ for all possibilities
  • $\frac{2!}{(2-2)!} = 2$ for two reds
  • $\frac{7!}{(7-1)!} = 7$ for first non-red
  • $\frac{2!}{(2-1)!} = 2$ for second red when first wasn't red.

If you put that into your equation you'll get exactly the same result as provided by others, ending up with the final $\frac29$ result.


Well, I stumbled across this and I have no idea why but I felt compelled to solve it and I used the following technique. There are exactly two cases where the second ball is red:

    1. The first ball is red, the second ball is red
    1. The first ball is not red, the second ball is red

For 1, the chances are (2/9)*(1/8)

For 2 the chances are (7/9)*(2/8)

Multiplying the probabilities you get 144/5184 and 1008/5184. Then you add these up to get 1152/5184. Finally, reducing the fraction 1152/5184 yields 2/9.

  • $\begingroup$ Of course now that I wrote out the answer and looked at it, very obviously the outcome of the first ball is irrelevant... $\endgroup$ – Vinnie Falco Jul 25 '18 at 14:08
  • $\begingroup$ Ah, I just realized why I wasted my time with this - procrastination of my other tasks. $\endgroup$ – Vinnie Falco Jul 25 '18 at 14:09

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