boredsatan
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Can someone please verify if my answers are correct for the below questions?
People have one of four different blood types – O, A, B or AB. Suppose 49% of the population have blood type O, 38% have type A, 10% have type B and 3% have type AB.
a) What is the probability that a randomly selected person
i) does not have type O blood? 0.51
ii) has type A or type B blood? 0.48
iii) is neither type O nor type A? 0.13
b) Among 4 random patients, what is the probability that
i) all have type O? (0.49)^4 = 0.05764801
ii) none of them are type A? (0.62)^4 = 0.14776336
iii) at least one person is type B? 1 - ((0.90)^4) = 0.3439
iv) the third patient only is type A? 0.62*0.62*0.38*0.62 = 0.09056464
Suppose a new cancer treatment has a 20% chance of curing a patient
a) For 5 cancer patients, what is the probability that:
i) none will be cured? 5C0 * (0.2)^0 * (0.8)^5 = 0.32768
ii) exactly 2 patients will be cured? 5C2 * (0.2)^2 * (0.8)^3
iii) at least 2 patients will be cured? P(X = 1) = 5C1 * (0.2)^1 * (0.8)^4 = 0.4096
P(X = 0) = 0.32768
P(X >=2) = 1 - (0.32768 + 0.4096) = 0.26272
b) For 500 patients
i) how many would you expect to be cured by the treatment? 500 * 0.20 = 100 patients
ii) what is the probability that more than 75 of the patients are cured?
variance = np(1-p) = (500*0.20)(0.8) = 80
so the standard deviation = sqrt(80)
using normal distribution with mean as 100 and standard deviation as sqrt(80)
P(X > 75) = 0.9974
Thanks!
People have one of four different blood types – O, A, B or AB. Suppose 49% of the population have blood type O, 38% have type A, 10% have type B and 3% have type AB.
a) What is the probability that a randomly selected person
i) does not have type O blood? 0.51
ii) has type A or type B blood? 0.48
iii) is neither type O nor type A? 0.13
b) Among 4 random patients, what is the probability that
i) all have type O? (0.49)^4 = 0.05764801
ii) none of them are type A? (0.62)^4 = 0.14776336
iii) at least one person is type B? 1 - ((0.90)^4) = 0.3439
iv) the third patient only is type A? 0.62*0.62*0.38*0.62 = 0.09056464
Suppose a new cancer treatment has a 20% chance of curing a patient
a) For 5 cancer patients, what is the probability that:
i) none will be cured? 5C0 * (0.2)^0 * (0.8)^5 = 0.32768
ii) exactly 2 patients will be cured? 5C2 * (0.2)^2 * (0.8)^3
iii) at least 2 patients will be cured? P(X = 1) = 5C1 * (0.2)^1 * (0.8)^4 = 0.4096
P(X = 0) = 0.32768
P(X >=2) = 1 - (0.32768 + 0.4096) = 0.26272
b) For 500 patients
i) how many would you expect to be cured by the treatment? 500 * 0.20 = 100 patients
ii) what is the probability that more than 75 of the patients are cured?
variance = np(1-p) = (500*0.20)(0.8) = 80
so the standard deviation = sqrt(80)
using normal distribution with mean as 100 and standard deviation as sqrt(80)
P(X > 75) = 0.9974
Thanks!