# Probability of a picking a white ball in second draw

Qn:Two balls are selected sequentially from an urn containing six red, three white, and four blue balls. What is the probability of selecting a white ball on the second draw if the first ball is not replaced before the second is selected?

My attempt:

The first ball can be white or non-white(red or blue).

case 1:The first ball is white: Then the probability of second ball is white is $3/13*2/12 = 1/26$.

case 2:The first ball is non-white: Then the probability of second ball is white is $10/13*3/12 = 5/26$.

So the total probability is $1/26 + 5/26 = 3/13$. Is this correct?

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Yes.

A more rigorous way of writing it would be using conditioning.

Let $W_i$ be the random variable corresponding to drawing of white balls.

$$W_i=\begin{cases}1\text{ If ball i is white}\\0\text{ Otherwise}\end{cases}$$

\begin{align} P(W_2=1)&=\sum_xP(W_2=1\;|\;W_1=x)\times P(W_1=x)\\ &=P(W_2=1\;|\;W_1=0)\times P(W_1=0)+P(W_2=1\;|\;W_1=1)\times P(W_1=1)\\ &=\dfrac{10}{13}\times \dfrac{3}{12}+\dfrac{3}{13}\times \dfrac{2}{12}\\ &=\dfrac{3}{13}\\ &\blacksquare \end{align}

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Why you added these probabilities because these are mutually exclusive? – Vinod Mar 3 '13 at 12:36
@Vinod. Implicitly, yes. I used the law of total probability. So, I condition on disjoint events and sum up the conditional probabilities. As you may have noticed in many problems, conditioning on something useful makes the problem really simple. – Inquest Mar 3 '13 at 12:38

Yes, this is indeed a correct answer

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