Thread: Need help with conditional probability and independent events problem!

1.If A and B are two events, find P(A and B) if
(a) P(A) = 0.5, P(B) = 0.7, P(A/B) = 0.4
(b) P(not A) = 0.3, P(not B) =0.5 and P(A/B) = 0.6

2.In a particular geographic region, 60 percent of the people are regular smokers, and evidence indicates that 10 percent of the smokers have lung cancer. Find the percent of the people who sre smokers and have lung cancer in this region.

3. A petroleum company exploring for oil has decided to drill two wells, one after the other. The probability of striking oil in the first well is 0.2. Given that the first attempt is successful, the probability of striking oil on the second attempt is 0.8. what is the probability of striking oil on both wells?

Hello,

Originally Posted by divinewisdom0218

1.If A and B are two events, find P(A and B) if
(a) P(A) = 0.5, P(B) = 0.7, P(A/B) = 0.4
(b) P(not A) = 0.3, P(not B) =0.5 and P(A/B) = 0.6

Conditional probability formula :


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2.In a particular geographic region, 60 percent of the people are regular smokers, and evidence indicates that 10 percent of the smokers have lung cancer. Find the percent of the people who sre smokers and have lung cancer in this region.

Percentages can be assimilated to "probability of finding a ... if we pick up an inhabitant".

P(smoker)=60%

P(lung cancer / smoker)=10% << that's because it is said that 10% of the smokers have lung cancer.

Same formula as above :
P(smoker and lung cancer)=P(lung cancer / smoker) * P(smoker)

3. A petroleum company exploring for oil has decided to drill two wells, one after the other. The probability of striking oil in the first well is 0.2. Given that the first attempt is successful, the probability of striking oil on the second attempt is 0.8. what is the probability of striking oil on both wells?

P(striking oil in the first well)=0.2
P(striking oil on the second attempt / stroke oil in the first well)=0.8

use the same formula