Question

Knowing that 80% of the volunteers donating blood in a clinic have type A blood.

(a) If five volunteers are randomly selected, what is the probability that at least one does not have type A blood?

(b) If five volunteers are randomly selected, what is the probability that at most four have type A blood?

(c) What is the smallest number of volunteers who must be selected if we want to be at least 90% certain that we obtain at least five donors with type A blood?

 

 

Answer 1

Solution

(a) If five volunteers are randomly selected, what is the probability that at least one does not have type A blood?

Let X : number of volunteer who has type A

\( \implies P(5-X\geq 1)=P(X\leq 4)=1-P(X=5) \)

we have  \( P(X=x)=C_n^xP^x\left(1-P\right)^{n-x}\hspace{3mm},X\sim Bin(5,0.8) \)

\( \implies P(X=5)=(0.8)^5=0.32768 \)

Therefore.  \( P(X\leq 4)=0.67232 \)

(b) Probability that at most four have type A. 

\( \implies P(X\leq 4)=0.67232 \)

(c) what is the smalle st number of volunteers 

If  \( P(X\geq 5)\geq 90\%=0.9 \)

we have  \( P(X\geq 5)=P(X=5)+...+P(X=n) \)

For For\( \hspace{2mm} n=5\implies P(X=5)=0.32768 \)

\( n=6\implies P(X\geq 5)=P(X=5)+P(X=6)=0.65536 \)

\( n=7\implies P(X\geq5)=0.852 \)

\( n=8\implies P(X\geq5)=0.9437 \)

Therefore.\( n=8 \)

Therefore.

a). \( P(X\leq 4)=0.67232 \)

b). \( P(X\leq 4)=0.67232 \)

c). \( n=8 \)