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%