Description
eResources | Chapter | Exercise | Page* | PDF Page* |
Lane et al. | 5 | 5 | 216 | 216 |
7 | 216 | 216 | ||
9 | 217 | 217 | ||
27 | 221 | 221 | ||
Illowsky et al. | 3 | 86 | 220 | 228 |
98 | 223 | 231 | ||
112 | 225 | 233 | ||
4 | 72 | 286 | 294 | |
80 | 289 | 297 | ||
88 | 290 | 298 |
1.Introduction to Statistics, David M. Lane et al., 2013. (http://onlinestatbook.com/Online_Statistics_Education.pdf)
2.Introductory Statistics, Barbara Illowsky et al., 2013.
(https://d3bxy9euw4e147.cloudfront.net/oscms-prodcms/media/documents/Statistics-OP.pdf)
Explanation & Answer
The solution is attached.I would gladly answer any question you have regarding the solution.
Introduction to Statistics, David M. Lane et al., 2013.
Question 5
A fair coin is flipped 9 times. What is the probability of exactly 6 heads?
Solution
Probability(X = x successes) =
𝑛
𝐶𝑥 ∙ 𝑝 𝑥 ∙ (1 − 𝑝)𝑛−𝑥
Where:
success is defined as outcome of a head
number of trials, n = 10
1
1
probability of success, p = 2
1 6 1 9−6
Probability(X = 6 heads) = 𝐶6 ∙ ( ) ∙ ( )
=
2
2
9
1 6 1 3
1 1
Probability(X = 6 heads) = 𝐶6 ∙ ( ) ∙ ( ) = 84 ×
×
2
2
64 8
9
Probability(X = 6 heads) =
1
probability of failure = 1 − p = 1 − 2 = 2
𝟐𝟏
𝟏𝟐𝟖
Therefore, the probability of exactly 6 heads in 9 flips of a coin is
𝟐𝟏
𝟏𝟐𝟖
Question 7
You flip a coin three times.
(a) What is the probability of getting heads on only one of the flips.
(b) What is the probability of getting head on at least one flip.
Solution
The sample space of flipping a coin three times is:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
(a) Probability(getting only one head in 3 flips) =
Number of outcomes with only one head
Total Number of outcomes
Outcomes with only one head = HTT, THT and TTH = 3 outcomes
∴ Pr(getting only one head in 3 flips) =
𝟑
𝟖
(b) Probability of getting head on at least one flip
Pr(at least one head) =
Number of outcomes with at least one head 𝟕
=
Total Number of outcomes
𝟖
Question 9
A jar contains 10 blue marbles, 5 red marbles, 4 green marbles, and 1 yellow marble. Two
marble are chosen (without replacement).
(a) What is the probability that one will be green and the other red?
(b) What is the probability that one will be blue and the other yellow?
Solution
Given:
Number of blue, B marbles = 10
Number of red, R marbles = 5
Number of green, G marbles = 4
Number of yellow, Y marbles = 1
Total number of marbles = 10 + 5 + 4 + 1 = 20
If the marbles are selected without replacement,
(a) Probability that one would be green and the other red is:
Pr(One green and other red) = Pr(first G and second R) + Pr(first R and second G)
4
5
5
4
1
1
Pr(One green and other red) = ( × ) + ( × ) =
+
20 19
20 19
19 19
𝟐
Pr(One green and other red) =
𝟏𝟗
(b) Probability that one would be blue and the other yellow is:
Pr(One blue and other yellow) = Pr(BY) + Pr(YB)
10 1
1 10
2
2
Pr(One blue and other yellow) = ( × ) + ( × ) =
+
20 19
20 19
19 19
𝟒
Pr(One blue and other yellow) =
𝟏𝟗
Question 27
A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2
mangos.
(a)...