A bag contains 13 white balls and 11 black balls. If two balls are drawn consecutively without replacement, what is the probability that both balls are white?
One way of answering this question is to draw a tree diagram. As there are only two colored balls, the first ball could be white or black and the second ball could be white or black. This means that we have four possible combinations. We could draw two white balls, a white then a black ball, a black then a white ball, or two black balls.
In this question, we need to calculate the probability that both balls are white. This means we are only interested in the top path, two white balls. As there were 13 white balls and 11 black balls, there were 24 in total. The probability that the first ball is white is 13 out of 24. As the ball is not being replaced, on the second draw there are 23 balls in the bag. As the first one selected was white, the probability that the second ball is also white is 12 out of 23 as there are 12 white balls left and 23 in total.
We recall that the AND rule in probability means multiply. In this case, we want to work out the probability that the first ball was white and the second ball was white. This means that we need to multiply 13 over 24 by 12 over 23. 12 and 24 are both divisible by 12. So we can cross cancel. Multiplying the numerators 13 and one gives us 13. And multiplying the denominators gives us 46. The probability that both balls that are drawn are white is 13 out of 46.
Whilst it is not required in this question, we could complete the tree diagram as shown. We could then use these probabilities to calculate the probability of drawing a white then a black, a black then a white, or two black balls. These are equal to 143 out of 552, 143 out of 552 again, and then 55 out of 276, respectively.