# Video: Pack 3 • Paper 3 • Question 12

Pack 3 • Paper 3 • Question 12

03:03

### Video Transcript

Lisa has a bag of red and blue marbles. There are 40 marbles in total, of which 23 are red. Two marbles are chosen at random. Lisa creates a probability tree to help her record this information. Write down one mistake you can see in this probability tree diagram.

The probabilities for the first marble are correct: there are 23 red marbles and 17 blue marbles. Therefore, the probability that the first marble is red is 23 out of 40 and the probability the first marble is blue is 17 out of 40.

When selecting the second marble, there are 39 of them left in the bag. This means that the denominator for the four fractions for the second marble must all be 39. We can also prove that this part of the diagram is incorrect by adding the two fractions circled. Each pair of branches must add up to one. Twenty-two fortieths plus seventeen fortieths equals thirty-nine fortieths. This is not equal to one. Likewise, twenty-three fortieths plus sixteen fortieths also equals thirty-nine fortieths, which again is not equal to one.

One mistake on Lisa’s probability tree diagram is that the probabilities for the second marble do not add up to one.

The second part of this question says the following.

Joe and Diana are both competing in an archery competition tomorrow. The probability that Joe will hit a bullseye during the competition is 0.45. The probability that Diana will hit a bullseye during the competition is 0.35. The archery coach says, “the probability that both Joe and Diana will hit a bullseye during the competition is 0.45 plus 0.35.” Is the archery coach correct? Give a reason for your answer.

The key word in this question is “and.” The archery coach is saying that both Joe and Diana will hit a bullseye. Well, the AND rule in probability means multiply, whereas the OR rule means add or plus. The archery coach is incorrect. The correct calculation is 0.45 multiplied by 0.35 and not 0.45 plus 0.35 as they suggested.

The probability that both Joe and Diana hit a bullseye tomorrow is 0.45 multiplied by 0.35.