#### boredsatan

##### Member

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- Mar 23, 2017

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- 1998

**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

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

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!