Question

Only about 75% of all donated human blood can be used in hospitals. The remaining 25% cannot be used because of various infections in the blood. Suppose a blood bank has 10 newly donated pints of blood. Let r be a binomial random variable that represents the number of "good" pints that can be used.
(a) Based on questionnaires completed by the donors, it is believed that at least 6 of the 10 pints are usable. What is the probability that at least 8 of the pints are usable, given this belief is true? Compute P(8 ≤ r | 6 ≤ r). (Round your answer to three decimal places.)
(b) Assuming the belief that at least 6 of the pints are usable is true, what is the probability that all 10 pints can be used? Compute P(r = 10 | 6 ≤ r). (Round your answer to three decimal places.)

Answer 1

Answer:

a) 0.02739

b) 0.00034

Step-by-step explanation:

If it is believed that at least 6 of the 10 pints are usable, then the 75% of “good” pints that can be used from this sample reduces to 60% of 75% = 0.6*0.75 = 0.45

Now for this sample we use a binomial distribution with probability of “success” (finding a “good” pint) of 0.45% and

the probability of getting exactly r good pints out of 10 is  

$\bf \binom{10}{r}0.45^r(1-0.45)^{10-r}=\binom{10}{r}0.45^r0.55^{10-r}$

where

$\bf \binom{10}{r}$ are combinations of 10 taken r at a time.

a)

The probability that at least 8 of the pints are usable is P(r>7)

$\bf \binom{10}{8}0.45^80.55^2+\binom{10}{9}0.45^90.55^1+\binom{10}{10}0.45^{10}0.55^0=\\0.02289+0.00416+0.00034=0.02739$

b)

Here we want P(r=10)

$\bf \binom{10}{10}0.45^{10}0.55^0=0.00034$