# Video: Pack 4 • Paper 3 • Question 10

Pack 4 • Paper 3 • Question 10

04:05

### Video Transcript

Andrew is taking part in an archery competition. In each round, he shoots at target A and then at target B. The probability that he hits target A is 0.7. The probability that he hits target B is 0.6. This is shown in the tree diagram. Andrew takes part in a number of rounds. In 21 of the rounds, he hits both target A and target B. Estimate the number of rounds in which he hits exactly one of the targets.

Let’s start by calculating the probability of each of the outcomes from each round. These are a hit and then a hit, a hit and then a miss, a miss then a hit, and a miss and a miss. Remember for two independent events A and B, the probability of both A and B occurring is equal to the probability of A multiplied by the probability of B. This is sometimes informally called the AND rule.

We can work out the probability that Andrew hits both target A and target B by multiplying these two probabilities. 0.7 multiplied by 0.6 is 0.42. Similarly, the probability that he hits target A and then misses target B is given by 0.7 multiplied by 0.4, which is 0.28. The probability that he misses target A and then hits target B is 0.3 multiplied by 0.6, which is 0.18. And the probability he misses both targets is 0.3 multiplied by 0.4, which is 0.12.

At this point, we can perform a quick check of our answer so far. The sum of all probabilities of all the outcomes should be one. 0.42 add 0.28 add 0.18 add 0.12 equals one. So our calculations so far are correct.

Next, we need to use the information that he hits both targets in 21 of the rounds. This will help us calculate the total number of rounds. Let’s call that 𝑥 for now. Remember if we knew the number of rounds that Andrew had completed, we could work out the number of rounds we would expect him to hit both targets in by multiplying this by the probability. 𝑥 is the total number of rounds and the probability that he hits both targets is 0.42. We actually do know that this happened 21 times. So we can form a linear equation in terms of 𝑥. To solve this equation, we’ll start by dividing both sides by 0.42. 21 divided by 0.42 is 50. This means the total number of rounds that Andrew took part in was 50.

Now, we want to calculate an estimate for the number of rounds during which he hit exactly one target. In this case, that’s the probability of him hitting target A and missing target B or missing target A and hitting target B. We can calculate that probability by adding together 0.28 and 0.18 to give us 0.46. The probability he hits exactly one target is, therefore, 0.46.

Now, remember we can calculate an estimate for the number of rounds in which this might happen by multiplying the probability by the total number of rounds. The probability is 0.46 and the number of rounds is 50. 0.46 multiplied by 50 is 23. The estimate for the number of rounds in which he hit exactly one of the targets is, therefore, 23.