(b) Work out the probability Jennifer hits the Bullseye at least once. Can someone work this out for me ?

1/2

Step-by-step explanation:

Filling out the tree diagram,

On the first throw, there is a 4/4 = 100%* chance Jennifer hits or misses. Therefore, the chance she misses is equal to the difference between

(chance she hits or misses) - (chance she hits) = 4/4 - 1/4 = 3/4

For the second throw, the chance she hits is independent of whether she hits or misses on the first throw. This means that it does not matter whether she hits or misses on the first throw; her chance to hit the second throw will always be 1/3. Therefore, 1/3 can go into the hit category on each hit line on the second throw, and in the miss category, we can put

100% - 1/3 = 1 - 1/3 = 3/3 - 1/3 = 2/3

For Jennifer to hit the bullseye at least once, she must either:

- Hit and then hit

- Hit and then miss

- Miss and then hit

From this, we can determine that if Jennifer hits the first bullseye, she will have hit it at least once, no matter what. Therefore, Jennifer must either:

- Hit the first one

- Miss and then hit

In probability, OR /either means that we add the probabilties. The probability that she hits the first one is 1/4, and the probability that she misses and then hits is

(probability she misses the 1st one) * (probability she hits the second one). We know to multiply here because both scenarios must happen, and for AND probabilities, we multiply. We thus have

(3/4) * (1/3) = 3/12 = 1/4 as the probabilty that she misses then hits, and 1/4 as the probability that she hits the first one. We add these up to get 1/4 + 1/4 = 2/4 = 1/2 as our probability that she hits the bullseye at least once

* we know 4/4 = 100% because 100% = 1, and anything divided by itself is equal to 1.

step-by-step explanation: where's the phrase?

because they could drop even lower..