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 high 13 10 content low 47 30 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)
b) P(B)
c) P (A|B)
d) P (BA)

Answer 1

Answer: a. 0.40   b. 0.23  c . 0.435   d . 0.25

Step-by-step explanation:

                                   melanin      content    Total

                                            high   low

moisture   high                     13      10                23

content    low                       47      30                77

 Total                                   60      40               100

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.

a) Total skin samples has low melanin content = 10+30=40

P(A)=$\dfrac{40}{100}=0.40$

b) Total skin samples has high moisture content = 13+10=23

P(B) =$\dfrac{23}{100}=0.23$

c) A ∩ B =  Total skin samples has both low melanin content and high moisture content =10

P(A ∩ B) =$\dfrac{10}{100}=0.10$

Using conditional probability formula , $P (A|B)=\dfrac{P(A\cap B)}{P(B)}$

$P (A|B)=\dfrac{0.10}{0.23}=0.434782608696\approx0.435$

d)  $P (B|A)=\dfrac{P(A\cap B)}{P(A)}$

$P (B|A)=\dfrac{0.10}{0.40}=0.25$