Video Transcript
A bag contains nine purple balls and six blue balls. If two balls are drawn one after another without replacement, find the probability that one of them is blue and the other is purple.
One way of answering a question of this type is to draw a tree diagram. We know that two balls are being selected and they could either be purple or blue. This means that four different combinations could occur. We could select two purple balls as shown by the top path. We could draw a purple followed by a blue or a blue followed by a purple. Finally, two blue balls could be drawn from the bag. As we want to find the probability that one ball is blue and the other is purple, we will be interested in the middle two paths, purple and blue or blue and purple.
We can now fill in the probabilities on the branches of the tree diagram. There are nine purple balls, and there are 15 balls in total. Therefore, the probability that the first ball is purple is nine out of 15. As there are six blue balls, the probability that the first ball is blue is six out of 15. These two fractions must be the compliment of each other; they must add to one.
As the ball is not being replaced, on the second draw, our probabilities will be out of 14. If the first ball drawn is purple, there will be eight purple balls left in the bag. Therefore, the probability the second ball will be purple is eight out of 14. The probability that the second ball is blue if the first ball is purple is six out of 14 as there are still six blue balls in the bag. However, if the first ball drawn was blue, the probability the second ball is purple would be nine out of 14. And the probability it would be blue is five out of 14.
We recall that when moving along our branches with tree diagrams, we need to multiply the probabilities. This is because we’re using the AND rule. To calculate the probability that the first ball is purple and the second is blue, we need to multiply nine over 15 by six over 14. Both of these fractions cancel, so we have three over five multiplied by three over seven. Multiplying the numerators gives us nine, and multiplying the denominators gives us 35. Therefore, the probability of drawing a purple ball then a blue ball is nine out of 35.
We can repeat this process for blue and then purple. This time, we’re multiplying six over 15 by nine over 14. Six over 15 cancels to two over five. We can also cross simplify or cross cancel so that we are left with one over five multiplied by nine over seven. Once again, this gives us an answer of nine over 35.
As there are two different ways of getting one blue ball and one purple ball, we have either purple and blue or blue and purple. We recall that the OR rule with probability means add. We need to add nine over 35 and nine over 35. As the denominators are the same, we just need to add the numerators, giving us 18 over 35. The probability when drawing two balls that one is blue and the other is purple is eighteen thirty-fifths or 18 out of 35.
