# Thread: Probabilities: Cars passing Tests

1. ## Probabilities: Cars passing Tests

hello;

I'm working on a probability example, if i know how to do this example, I will be able to do well in exam, hopefully someone can help me out...

Here is the scenario;

Before cars are brought to the market they under go three tests. There are 7200 cars, and here are the results for the tests;

• 288 fail at the first test
• 240 fail at the second test
• 144 fail at test three...

I have the following questions and my answers to them;

a. probability for passing the first test

I said: 1 - 288/7200 = 0.96%

b. probability of car passing the 2nd (not included the first test)

I said: 1 - 240 / 7200 = 0.96%

c. probability of car passing 3rd test by itself

I said: 1 - 144/7200 = 0.98%

d. probability of passing ALL the three tests

I said: 0.96 x 0.96 x 0.98 = 90%

e. probability of failing at test two

Can someone please give me some feedback on my answers, whether I'm doing it in the right way and also whether this is the only (easy) way of doing it...

Thanks!

2. (a) is right
(b,c,etc.) watch out - you no longer have 7200 total cars because some failed at the first test.

(e) failing at test 2: 240 / (7200-288)

3. Originally Posted by Student122
hello;

I'm working on a probability example, if i know how to do this example, I will be able to do well in exam, hopefully someone can help me out...

Here is the scenario;

Before cars are brought to the market they under go three tests. There are 7200 cars, and here are the results for the tests;

• 288 fail at the first test
• 240 fail at the second test
• 144 fail at test three...

I have the following questions and my answers to them;

a. probability for passing the first test

I said: 1 - 288/7200 = 0.96%

b. probability of car passing the 2nd (not included the first test)

I said: 1 - 240 / 7200 = 0.96%

c. probability of car passing 3rd test by itself

I said: 1 - 144/7200 = 0.98%

d. probability of passing ALL the three tests

I said: 0.96 x 0.96 x 0.98 = 90%

e. probability of failing at test two

Can someone please give me some feedback on my answers, whether I'm doing it in the right way and also whether this is the only (easy) way of doing it...

Thanks!
a) 6912 cars pass the first test, so the probability of passing is 6912/7200. This is the same as your answer, just worked a slightly different way.

b) 6912 cars are tested and 6672 pass, so the probability of passing is 6672/6912.

c) 6672 cars are tested and 6528 pass, so ...

d) 6528 cars out of 7200 pass all 3 tests, so ....

e) 240 out of 7200 cars fail at test 2, so ...