1. ## Probability Word Problems

Can anyone help me understand how to solve these word problems? they are so confusing. There are 3 in total:

1) An advertising agency notes that approximately 1 in 50 potential buyers of a product sees a given magazine ad and 1 in 5 sees a corresponding ad on television. 1 in 100 sees both. 1 in 3 actually purchases the product after seeing the ad, 1 in 10 without seeing it. What is the prbability that a randomly selected potential customer will purchase the product?

2)In a given region, the pobability of finding oil is 0.1. A test is often used to test for oil prior to drilling. From past experience it is known that when there is oil, the test reads positive 95% of the time and when there is not oil the test reads negative 80% of the time. If you perform the test and it reads positive, what is the probability that there is oil?

3)A study of residents of a region showed that 20% were smokers. The probabilityof death due to lung cancer, given that a person smoked was ten times the probability of death due to lung cancer given that they did not smoke. If the probability of death due to lung cancer in the region is .006, what is the probability of death due to lung cancer given that that person is a smoker?

Any help you guys can give me would be greatly appreciated. These are the only three questions on my final that im going to fail!!! I really cant organize these ideas in my head and turn them into math problems. Thanks in advance!

2. Originally Posted by aelunyu
Can anyone help me understand how to solve these word problems? they are so confusing. There are 3 in total:

1) An advertising agency notes that approximately 1 in 50 potential buyers of a product sees a given magazine ad and 1 in 5 sees a corresponding ad on television. 1 in 100 sees both. 1 in 3 actually purchases the product after seeing the ad, 1 in 10 without seeing it. What is the prbability that a randomly selected potential customer will purchase the product?
Suppose there were 1000 potential customers, then (on avaerge) 20 will have seen
the magazine add, and 200 will have seen the TV add and 10 will have seen
both. So 20+200-10=210 will have seen at least one of the adds, and 790
will have seen neither.

So our sales are 210/3 + 790/10 = 149 out of 1000 potential customers.
Hence th probability that a random customer buys is 149/1000 = 0.149
or 14.9%.

RonL

3. Originally Posted by aelunyu
2)In a given region, the pobability of finding oil is 0.1. A test is often used to test for oil prior to drilling. From past experience it is known that when there is oil, the test reads positive 95% of the time and when there is not oil the test reads negative 80% of the time. If you perform the test and it reads positive, what is the probability that there is oil?
Again imagine 1000 potential wells, of these on avaerage 100 will contain oil
and of these 95 will test positive. Of the remaining 900 with no oil 180 will
test positive (20% of those with no oil test positive). So there are a total
of 280 which test positive of which 100 contain oil, so the required probability
is:

p = 100/280 ~=0.357