32% of U. S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five, inclusive

A.  21.06%

B. 66.88%

C. 81.56%

Step-by-step explanation:

There are two possible answer for the question asked, yes(more likely to make purchases) or no. The probability for a random adults answer yes to the question is 32% so the probability that the adults answer no is 68%.

(a) exactly two

There is only one case for this question, 2 people say yes and 8 people say no. The calculation for this problem will be:

2P10 * P(yes)^2 * P(no)^8= 10!/2!8!* 32%^2  * 68%^8 = 0.21066= 21.06%

(b) more than two

There are a lot of case for more than two, its easier to find out the negation of the probability which is "two or less". Case that fulfill "two or less" will be: 2 yes and 8 no

1 yes and 9 no

0 yes and 10 no

The calculation for the negation will be:

~P(X)= 0P10 * P(yes)^0 * P(no)^10   +   1P10 * P(yes)^1 * P(no)^9   +  2P10 * P(yes)^2 * P(no)^8  =

10!/0!10!* 32%^0  * 68%^10    +   10!/1!9!* 32%^1  * 68%^9   +   10!/2!8!* 32%^2  * 68%^8

0.02113 + 0.09947 + 0.21066  =  0.33126= 33.12%

Since its the negation, you need to subtract 1 with the negation

P(X)= 1 - ~P(X)

P(X)= 1 - 33.12%= 66.88%

(c) between two and five, inclusive

Same formula as above, but the case is:

2 yes and 8 no

3 yes and 7 no

4 yes and 6 no

5 yes and 5 no

The calculation will be:

2P10 * P(yes)^2 * P(no)^8 + 3P10 * P(yes)^3 * P(no)^7 + 4P10 * P(yes)^4 * P(no)^6 + 5P10 * P(yes)^5 * P(no)^5 =

10!/2!8!* 32%^2  * 68%^8    +   10!/3!7!* 32%^3  * 68%^7   +   10!/4!6!* 32%^4  * 68%^6  +  10!/5!5!* 32%^5  * 68%^5

0.21066 + 0.26435 + 0.21770 + 0.1229= 0.81561= 81.56%

i really dont know

step-by-step explanation: