Question

Samples of skin experiencing desquamation are analyzed for both moisture and melanin content. The results from 100 skin samples are as follows:

		melanin content
		high	low
moisture content	high	13	9
	low	45	33

Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high moisture content. Determine the following probabilities. Round your answers to three decimal places (e.g. 98.765).

a) P(A)= Enter your answer in accordance to the item a) of the question statement
b) P(B)= Enter your answer in accordance to the item b) of the question statement
c) P(A|B)= Enter your answer in accordance to the item c) of the question statement
d) P(B|A)= Enter your answer in accordance to the item d) of the question statement

Answer 1

The results are given below :

	Melanin Content	Total
High	Low
Moisture Content	High	13	9	22
Low	45	33	78
Total	58	42	100

Let A denote the event a sample has low melanin content.

Let B denote the event that a sample has high moisture content.

a) Answer :

P(A) = number of samples that have low melanin content / total number of samples

        = 42 / 100

        = 0.420

Therefore, P(A) = 0.420

b) Answer :

P(B) = number of samples that have high moisture content / total number of samples

        = 22 / 100

        = 0.220

Therefore, P(B) = 0.220

c) Answer :

P(A | B) = P(A B) / P(B)     ...........(conditional probability)

Now, P(A B) = number of samples who have low melanin content and high moisture content / total number of samples

   = 9 / 100

   = 0.09

So, P(A | B) = 0.09 / 0.220

= 0.409

Therefore, P(A | B) = 0.409

d) Answer :

P(B | A) = P(A B) / P(A) ............(conditional probability)

= 0.09 / 0.420 ........(from parts c) and a))

= 0.214

Therefore, P(B | A) = 0.214