Question 1177125: 7. In Alberta, 55% of new drivers take a driver training course before completing their driver’s test. Of the drivers that took training, 75% pass their driver’s test. Of the drivers that did not take training, 58% pass their driver’s test.
a. Complete the following tree diagram to represent the given information. Express the probability of each event to 4 decimal places if necessary.
Diagram: -------> i= P(training)------>iii= P(pass)----> vii= P(training AND pass)
-------> iv= P(fail)----> viii= P(training AND fail)
-------> ii= P(no training)-----> v= P(pass)-----> ix=P(no training AND pass)
------> vi= P(fail)----> x= P(no training AND fail)
MY ANSWERS:
i. 55%
ii. 45%
iii. 75%
iv. 25%
v. 58%
vi. 42
vii. 0.42
viii. 0.14
ix. 0.26
x. 0.18
b. Jim passed his driver’s test. Use the values in your probability tree above to determine the probability that he took driver training, to the nearest whole percent? (1 mark)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! It looks like you've got a good handle on the tree diagram and probabilities! Here's a breakdown with some minor corrections:
**a. Completed Tree Diagram**
```
--- P(Pass | Training) = 0.75 --> P(Training AND Pass) = 0.55 * 0.75 = 0.4125
/
/
P(Training) = 0.55
\ --- P(Fail | Training) = 0.25 --> P(Training AND Fail) = 0.55 * 0.25 = 0.1375
\
/
--- P(Pass | No Training) = 0.58 --> P(No Training AND Pass) = 0.45 * 0.58 = 0.2610
\
/
P(No Training) = 0.45
\ --- P(Fail | No Training) = 0.42 --> P(No Training AND Fail) = 0.45 * 0.42 = 0.1890
\
/
```
**Minor Corrections:**
* **vii. P(Training AND Pass):** 0.55 * 0.75 = 0.4125
* **viii. P(Training AND Fail):** 0.55 * 0.25 = 0.1375
* **ix. P(No Training AND Pass):** 0.45 * 0.58 = 0.2610
* **x. P(No Training AND Fail):** 0.45 * 0.42 = 0.1890
You had the right idea with the calculations, just a slight adjustment in the values for vii and viii.
**b. Probability Jim Took Training Given He Passed**
We want to find P(Training | Pass). We can use Bayes' Theorem:
```
P(Training | Pass) = [P(Pass | Training) * P(Training)] / P(Pass)
```
To find P(Pass), we need to consider both ways someone can pass:
```
P(Pass) = P(Training AND Pass) + P(No Training AND Pass)
P(Pass) = 0.4125 + 0.2610 = 0.6735
```
Now we can plug the values into Bayes' Theorem:
```
P(Training | Pass) = (0.75 * 0.55) / 0.6735
≈ 0.6126
≈ 61% (to the nearest whole percent)
```
**Therefore, the probability that Jim took driver training given that he passed his test is approximately 61%.**
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