Question

One-third of blood donors at a clinic have O+ blood.

(a) Assuming that there is a very large number of donors at the clinic and the donors are independent of each other, find the probability that

(i) six or more donors have to be screened in order to find two who have O+ blood.
How many of the following discrete probabilities will be applicable (Answer should be one of: one, two, three)? (Binomial, geometric, negative binomial)

How to represent the event that 6 or more donors have to be screened in order to find two who have O+ blood? (

(A)  X ≤ 1 for X ∼ B i n ( 5 , 1 3 );

(B)  X ≤ 2 for X ∼ B i n ( 6 , 1 3 );

(C)  X ≥ 6 for X ∼ N e g B i n ( 2 , 1 3 );

(D)  X ≤ 2 for X ∼ N e g B i n ( 6 , 1 3 );

(E) Both (A) and (C)
)

(ii) among six donors screened, more than two do not have O+ blood.
How many of the following discrete probability distributions will be applicable? (one of: Binomial, geometric, negative binomial)

How to represent the event that there are more than two donors who do not have O+ blood among six donors screened?

(a)X≥3forX∼Bin(6,23)X≥3forX∼Bin(6,23)

(b)X≥2forX∼Bin(5,23)X≥2forX∼Bin(5,23)

(c)X=6forX∼Bin(2,23)X=6forX∼Bin(2,23)

(d)X=6forX∼NegBin(2,23)X=6forX∼NegBin(2,23)

(e)


(iii) more than six donors have to be screened in order to find the first with O+ blood.
How many discrete probabilities will be applicable? (one of: binomial, geometric, negative binomial)
How to represent the event that more than six donors have to be screened in order to find the first with O+ blood?

(A)  X = 6 f o r X ∼ B i n ( 6 , 2 3 );

(B)  X = 0 f o r X ∼ B i n ( 6 , 1 3 );

(C)  X > 6 f o r X ∼ N e g B i n ( 1 , 1 3 );

(D)  X > 6 f o r X ∼ G e o m ( 1 3 );

(E) All of the above.


(b) Find the expected number of donors who must be screened in order to find five with O+ blood.

Answer 1

i)

P(X=   2   )=   1C1*0.3333^2*0.667^0=       0.1111
P(X=   3   )=   2C1*0.3333^2*0.667^1=       0.1481
P(X=   4   )=   3C1*0.3333^2*0.667^2=       0.1481  
P(X=   5   )=   4C1*0.3333^2*0.667^3=       0.1317
P(6 or more) = 1 - 0.1111 - 0.1481- 0.1481 - 0.1317 = 0.5390

(C)  X ≥ 6 for X ∼ N e g B i n ( 2 , 1/3 );

ii) P(X>2) = 0.3196

(a)X≥3forX∼Bin(6,23)X≥3forX∼Bin(6,2/3)

iii)

P(x>6) = 0.0879

(D)  X > 6 f o r X ∼ G e o m ( 1/3 );

iv)

mean=µ = r/p= 5/(1/3) = 15