Answer:
The probability of getting someone who tests positive, given that he or she did not have the disease = P(P|No) = 0.041
Step-by-step explanation:
Complete Question
The data represents the results for a test for a certain disease. Assume one individual from the group is randomly selected. Find the probability of getting someone who tests positive, given that he or she did not have the disease.
P/N | Yes | No
+ve | 141 | 6
-ve | 11 | 142
Note that +ve and -ve l stands for testing positive and negative respectively.
Yes and No represents whether one has the disease or not respectively.
Solution
Let
- The event of testing positive be P.
- The event of testing negative be N.
- The event of having the disease be Yes.
- The event of not having the disease be No.
The probability of getting someone who tests positive, given that he or she did not have the disease = P(P|No)
The conditional probability of A given B is given mathematically as
P(A|B) = P(A n B) ÷ P(B)
Hence,
P(P|No) = P(P n No) ÷ P(No)
To solve these probabilities, we first define the probability of an event as the number of elements in that event divided by the Total number of elements in the sample space.
P(No) = n(No) ÷ n(Sample space)
n(No) = 6 + 142 = 148
n(Sample Space) = 141 + 11 + 6 + 142 = 300
P(No) = (148/300)
P(P n No) = n(P n No) ÷ n(Sample space)
n(P n No) = 6
n(Sample Space) = 300
P(P n No) = (6/300)
P(P|No) = P(P n No) ÷ P(No)
= (6/300) ÷ (148/300)
= (6/148)
= 0.0405405405 = 0.041 to 3 d.p.
Hope this Helps!!!
Explanation:
