Question

Consider 8 blood donors chosen randomly from a population. The probability that the donor has type A blood is .40. Which of the following is CORRECT?
(a) The probability that between 3 and 5 donors (inclusive) will have type A blood about .37.

(b) The probability of 1 or fewer donors having type A blood is about .11.

(c) The probability of 7 or more donors NOT having type A blood is about .0087.

(d) The probability of exactly 5 donors having type A blood is about .28.

(e) The probability of exactly 5 donors NOT having type A blood is about .12.

Answer 1

Answer: Option B is the only correct option.

Step-by-step explanation:

Number of samples = n = 8.

Probability of success = p = 0.4

Probability of failure = q = 0.6

r = chosen number of donors among the 8

To solve this question, we use the distribution formula

P(x=r) = nCr * p^r * q^n-r

For option A, to check if P(3<x<5) = 0.37. [3 and 5 inclusive]

When x = 3

P(x=3) = 8C3 * 0.4^3 * 0.6^5

P(x=3) = 56 * 0.064 * 0.07776

P(x=3) = 0.2787

When x= 4

P(x=4) = 8C4 * 0.4^4 * 0.6^4

P(x=4) = 70 * 0.0256 * 0.1296

P(x=4) = 0.2322

Since p(x=3) + p(x=4) is already greater than 0.37, then we know option A is NOT correct.

For option B, To check if the probability of 1 or fewer donor is about 0.11. i.e if P(x</=1) = 0.11

When x=o

P(x=0) = 8C0 * 0.4^0 * 0.6^8

P(x=0) = 1* 1 * 0.016796

P(x=0) = 0.016796.

When x = 1

P(x=1) = 8C1 * 0.4^1 * 0.6^7

P(x=1) = 8 * 0.4 * 0.02799

P(x=1) = 0.08958

P(x=0) + P(x=1) = 0.016796 + 0.08958

P(x=0) + P(x=1) = 0.10635.

Since this is approximately 0.11, then option B is a correct option.

For option C to check if the probability 7 or more donors not having type A = 0.0087

To do this,we determine thw probability of 7 or more donors having type A and we subtract our answer from 1.

First, we determine P(x>/=7)

When x= 7

P(x=7) = 8C7 * 0.4^7 * 0.6^1

P(x=7) = 8 * 0.001638 * 0.6

P(x=7) = 0.007864

When x=8

P(x=8) = 8C8 * 0.4^8 * 0.6^0

P(x=8) = 1 * 0.0006554 * 1

P(x=8) = 0.0006554

P(x=7) + P(x=8) = 0.007864 + 0.0006554 = 0.00852.

Since probability of 7 or more donors having type A is 0.00852 as against what was stated in the option C, then option C is NOT a correct option.

For option D, to check if the probability of exactly 5donors having type A blood = 0.28

When x=5

P(x=5) = 8C5 * 0.4^5 * 0.6^3

P(x=5) = 56 * 0.01024 * 0.216

P(x=5) = 0.1239.

Since probability of what was derived for having exactly 5 donors having sample A is different from what wqs given in the option, then option D is NOT correct.

For option E, since what was stated in the option negates what was derived for exactly 5 donors, then option E is NOT correct