Question

NO LINKS!!! URGENT HELP PLEASE!!!!

At her archery lessons, Tess has a 90% of getting a bull's eye. Unfortunately, there's only a 40% chance that her instructor will be watching her when she shoots. What is the probability that her instructor will see Tess his the bull's eye? Make a diagram and explain it completely.

Answer 1

0.36 or 36%.

Step-by-step explanation:

To solve this problem, we can use a tree diagram to visualize the different possible outcomes and their probabilities.

The first event is whether the instructor is watching or not, with a probability of 0.4 for the instructor watching and 0.6 for not watching. The second event is whether Tess hits the bull's eye or not, with a probability of 0.9 for hitting the bull's eye and 0.1 for missing it.

We can create a tree diagram with two branches, one for the instructor watching and one for the instructor not watching, and two branches coming out of each of those for Tess hitting the bull's eye or missing it. The probabilities for each branch can be written along the branches, as shown below:

Attachment 1: Tree Diagram for Tess's Archery Lessons

To find the probability that the instructor sees Tess hit the bull's eye, we need to look at the branch where the instructor is watching and Tess hits the bull's eye. This branch has a probability of 0.4 x 0.9 = 0.36. Therefore, the probability that the instructor sees Tess hit the bull's eye is 0.36 or 36%.

In summary, we can use a tree diagram to visualize the different outcomes and probabilities for Tess hitting the bull's eye and her instructor watching. From this, we can see that the probability of her instructor seeing her hit the bull's eye is 0.36 or 36%.

Answer 2

Answer: 36%

The diagrams are below.

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Step-by-step explanation:

There isn't a real need to use a diagram. We simply multiply the decimal values of each percentage given

• 90% ---> 0.90
• 40% ---> 0.40

There is a 0.90*0.40 = 0.36 = 36% chance of the instructor seeing Tess hit the bullseye.

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More formally,

A = Tess hits the bullseye

B = The instructor watches

P(A) = 0.90

P(B) = 0.40

P(A and B) = P(A)*P(B) since the events are independent

P(A and B) = 0.90*0.40

P(A and B) = 0.36

P(A and B) = 36%

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The diagrams are shown below if you still need them.

Figure 1 shows the probability tree. The values in red are the probability for each branch. The values in blue are the result of multiplying the red values along a certain pathway.

Example: Follow the left-most pathway to see 0.90*0.40 = 0.36 is the probability of "hits bullseye" and "instructor is watching" happening at the same time.

Figure 2 shows another way to view probability. We start with a 1 by 1 square. Its area is 1*1 = 1 square unit. This square is sub-divided into 4 regions marked A, B, C, D

• A = hits bullseye, instructor watches
• B = hits bullseye, instructor is not watching
• C = misses bullseye, instructor watches
• D = misses bullseye, instructor is not watching

Take note how region A is 0.90 units tall and 0.40 units across. Its area is 0.90*0.40 = 0.36; so this is a visual tool to see why multiplication takes place for problems like this. Regions C through D are a similar story.

We can arrange the results into a table like so

`\begin{array}c \cline{1-3}& \text{watching} & \text{not watching}\\\cline{1-3}\text{hit} & 0.36 & 0.54\\\cline{1-3}\text{miss} & 0.04 & 0.06\\\cline{1-3}\end{array}`

Notice that all four values add to 1.