Question

Figure 3.51 gives the estimated number of diagnoses of AIDS in the United States by age at the time of diagnosis in 2011. Also note that in 2011, there were 301,621,157 residents of the United States, according to the U.S. Census Bureau. Figure 51 AIDS by age. Cumulative Number of AIDS Cases 15 49 Age (Years) Under 13 13-14 15-24 25-34 35-44 2,935 7,434 8,874 8,561 3,305 899 45-54 55-64 65 or older Source: Centers for Disease Control and Prevention. a. Find the probability that a U.S. resident diagnosed with AIDS was 15 to 24 years old at the time of diagnosis. b. Find the probability that a U.S. resident was diagnosed with AIDS and was 15 to 24 years old at the time of diagnosis. c. Find the probability that a U.S. resident diagnosed with AIDS was 25 to 34 years old at the time of diagnosis. d. Find the probability that a U.S. resident was diagnosed with AIDS and was 25 to 34 years old at the time of diagnosis. e. Explain in words the difference between parts (c) and (d).

Answer 1

The total number of AIDs cases is

15 + 49 + 2935 +7434 + 8874-8561 + 3305 + 899 = 32072

(a)

Out of 32072 cases, 2935 was 15 to 24 years old at the time of diagnosis so the required probability is

P(15-24 years old | AIDs case) = 2935 / 32072 = 0.0915

(b)

The required probability is

P(15-24 years old and AIDs case) = 2935 / 301621157 = 0.00000973

(c)

P(25-34 years old | AIDs case) = 7434 / 32072 = 0.2318

(d)

P(25-34 years old and AIDs case) = 7434 / 301621157 = 0.0000246

(e)

The probability of part c is conditional probability while the probability of part d is joint probability.