If a bag has 8 balls in it 4 white and 4 black what are the odds against picking four black balls one after the other

First of all, I'm assuming you aren't returning any drawn ball to the bag. The probability against anything happening is simply one minus the probability for it happening. So if is the probability of picking four black balls, then the probability against drawing four black balls is just . So what we really need to know is the probability of picking four black balls. Since we start with eight balls total and four black balls, the probability of picking one black ball on our first draw is . Keep in mind that we can reduce this, but there is a pattern and in order for you to see it, I'll keep this fraction unreduced.

On the second drawing, assuming the first drawn ball is black, then we have seven total balls and only three black balls. So the probability of drawing a second black ball having drawn a black ball on the first drawing is . Similarly, the probability of drawing a third black ball, since there are still two black balls and six total balls, is . For the fourth drawing the probability is . So to find the probability of drawing four black balls we simply multiply the probability of drawing a black ball on the first drawing and the probability of drawing a second black ball having already drawn a black ball and the probability for the third, etc. So we get



Since the original problem was to find the probability against drawing four black balls, the answer you're looking for is



I hope this helps.

Just a point of clarification: Corrigan explained perfectly how to calculate the probability, which is the hard part. However, since the question asked about odds, not probability, you have to do one last small step and say the odds against are 69:1 (said as "69 to 1"), which is equivalent to a probability of 69/(69+1)=69/70 as Corrigan showed.

Oh, I didn't even notice that. Thanks jcarson2, stats has never been my strong point. :)