Diseases I and II are prevalent among people in a certain population. It is assumed that 11% of the population will contract disease I sometime during their lifetime, 16% will contract disease II eventually, and 2% will contract both diseases. (a) Find the probability that a randomly chosen person from this population will contract at least one disease. .25 Correct: Your answer is correct. (b) Find the conditional probability that a randomly chosen person from this population will contract both diseases, given that he or she has contracted at least one disease. (Round your answer to four decimal places.)

a)  

b)  

Step-by-step explanation:

From the question we are told that:

Probability of contacting disease 1

Probability of contacting disease 2

Probability of contacting both disease  

a)

Generally the equation for a Random contact is mathematically given by

b)

Generally the equation for Probability of contacting both after having contacted one is mathematically given by

Therefore

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt