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

Answer 2

The probability of 1 or fewer donors having type A blood is about .11 and this can be determined by using the distribution formula.

Given :

• Consider 8 blood donors chosen randomly from a population.
• The probability that the donor has type A blood is 0.40.

The distribution formula can be used to determine the correct statement. The distribution formula is given by:

$\rm P(x = r) = \; ^nC_r\times p^r \times q^{n-r}$

Now, check option wise:

A) P(3 $\leq$ x $\leq$ 5) = 0.37

$\rm P(x = 3) = \; ^8C_3\times 0.4^3 \times 0.6^{8-3}$

$\rm P(x = 3) = 56\times 0.064\times 0.07776$

P(x = 3) = 0.2787

$\rm P(x = 4) = \; ^8C_4\times 0.4^4 \times 0.6^{8-4}$

$\rm P(x = 4) = 70\times 0.0256\times 0.1296$

P(x = 4) = 0.2322

Before finding P(x = 5), add P(x = 3) and P(x = 4).

P(x = 3) + P(x = 4) = 0.2787 + 0.2322 = 0.5109

The above value is greater than 0.37 so this option is incorrect.

B) P(x $\leq$ 1) = 0.11

$\rm P(x = 0) = \; ^8C_0\times 0.4^0 \times 0.6^{8-0}$

$\rm P(x = 0) = 1\times 1\times 0.016796$

P(x = 0) = 0.016796

$\rm P(x = 1) = \; ^8C_1\times 0.4^1 \times 0.6^{8-1}$

$\rm P(x = 1) = 8\times 0.4\times 0.02799$

P(x = 1) = 0.8958

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

the above value is approximately 0.11 so this option is correct.

C) P(x $\geq$ 7) = 0.0087

$\rm P(x = 7) = \; ^8C_7\times 0.4^7 \times 0.6^{8-7}$

$\rm P(x = 7) = 8\times 0.001638\times 0.6$

P(x = 7) = 0.007864

$\rm P(x = 8) = \; ^8C_8\times 0.4^8 \times 0.6^{8-8}$

$\rm P(x = 8) = 1\times0.0006554\times 1$

P(x = 8) = 0.0006554

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

The probability of 7 or more donors having type A is 0.00852 therefore, it is incorrect option.

D) P(x = 5) = 0.28

$\rm P(x = 5) = \; ^8C_5\times 0.4^5 \times 0.6^{8-5}$

$\rm P(x = 5) = 56\times0.01024\times 0216$

P(x = 5) = 0.1239

Therefore, this option and option E) is incorrect.

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https://brainly.com/question/23017717