Question 980063
I'll do the first two parts (a and b) to get you started.


For parts a) through d), the following is true

<ul>
<li>n = 6 is the sample size</li>
<li>p = 0.40 is the probability of having the disease</li>
</ul>

The binomial probability distribution fromula will be used. The formula is P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k). I'm using the <a href="http://www.mathwords.com/c/combination_formula.htm">combination formula</a> which is (n C r) = (n!)/(r!*(n-r)!) for the "n C k" portion.


In part a), the value of k is k = 1. In part b), the value of k is k = 2.

=====================================================================

a)

Exactly One


k = 1


Binomial distribution


P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(x = 1) = (6 C 1)*(0.40)^(1)*(1-0.40)^(6-1)
P(x = 1) = (6)*(0.40)^(1)*(0.6)^(5)
P(x = 1) = 0.186624


Answer: <font color="red">0.186624</font>


=====================================================================

b)

Exactly Two


k = 2


Binomial distribution


P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(x = 2) = (6 C 2)*(0.40)^(2)*(1-0.40)^(6-2)
P(x = 2) = (15)*(0.40)^(2)*(0.6)^(4)
P(x = 2) = 0.31104


Answer: <font color="red">0.31104</font>


=====================================================================