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A bag contains 30 balls 20 red and 10 blue. Two balls are drawn from the bag. Let A be the event that the first ball is red, and B be the event that the second ball is red. Are A and B independent?

I have no idea how to solve this problem.

for A we want red for first position (20/30) and for second position it can be any color but for B if the first position comes out to be blue then we will have (20/30) but if we have red in first position ...

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One question to ask is whether the outcome of the first event affects the probability of the second event? – jimmywho Dec 4 '12 at 6:41
Yes, also if two events are independent P($A\bigcap B$)=P(A)$\bigcap$P(B). So if they are not equal then the two events can not be independent. – jimmywho Dec 4 '12 at 6:46
Can you please explain more? – Node.JS Dec 4 '12 at 6:52
up vote 4 down vote accepted

Presumably we are sampling without replacement.

Let $A$ be the event the first ball is red, and $B$ the event the second ball is red. Intuitively the answer is clear. Take an extreme example of $1$ red ball and $10$ blue. If we know the first ball was red, then we know for sure that the second will be blue. But we will do some formal calculations, since all too often in probability the intuition is not entirely reliable.

The probability that the first ball is red is $\dfrac{20}{30}$. Given that the first ball is red, the probability the second is red is $\dfrac{19}{29}$, for one red ball is drawn. So, in symbols, $\Pr(B|A)=\dfrac{19}{29}$.

Now what is the plain $\Pr(B)$? Imagine that the balls all have unique ID numbers, and that we keep drawing them out one at a time until all the balls have been drawn. All sequences of ID numbers are equally likely. Since there are $20$ red balls, the probability a red ball occupies the second position is simply $\dfrac{20}{30}$. This is $\Pr(B)$.

Since $\Pr(B|A)\ne \Pr(B)$, the events $A$ and $B$ are not independent.

You probably had another definition of independence, namely "the events $A$ and $B$ are independent if and only if $\Pr(A\cap B)=\Pr(A)\Pr(B)$."

We can check the non-independence by using the definition. By the discussion above, $\Pr(A\cap B)=\dfrac{20}{30}\cdot\dfrac{19}{29}$.

However $\Pr(A)\Pr(B)=\dfrac{20}{30}\cdot\dfrac{20}{30}$.

Remark: After a while, it will be obvious that given no information on the colour of the first ball, the probability that the second ball is red is $\dfrac{20}{30}$. But let's show this an uglier way.

The second ball can be red in two ways: (i) First is red and second is red or (ii) First is blue and second is red. By earlier discussion, the probability of (i) is $\dfrac{20}{30}\cdot\dfrac{19}{29}$.

By a similar argument, the probability of (ii) is $\dfrac{10}{30}\cdot\dfrac{20}{29}$.

Add up and simplify. After a while we get $\dfrac{2}{3}$, or equivalently $\dfrac{20}{30}$.

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As Always, very clear answer and really easy to understand, Thanks Man, Thanks – Node.JS Dec 4 '12 at 7:24
There's a little typo : "Since $\Pr(B|A)\ne \Pr(B)$,..." – Thibaut Dumont Dec 4 '12 at 8:34
@ThibautDumont: Thank you, fixed I hope. Unfortunately probably not the only one. – André Nicolas Dec 4 '12 at 8:45

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