AP Statistics / Mr. Hansen |
Name: __________KEY___________ |
1. |
Let A = the event of getting an even outcome on a die roll, B = the event of getting two heads when two fair coins are tossed, and C = the event of getting a 5 on a die roll. |
(a) |
The probability of getting two heads, given that the die roll is even = __1/4___ . Are A and B independent? _yes_ Why or why not? __P(B | A) = P(B) = 1/4____ |
(b) |
The probability of getting an even die roll and two heads = __1/8___ . Are A and B disjoint? __no_ Why or why not? ____P(A Ç B) = 1/8 ¹ 0______ |
(c) |
The probability of getting an even die roll or two heads = _1/2 + 1/4 – 1/8 = 5/8_ . |
(d) |
The probability of getting an even die roll or a 5 = __2/3__ . Are A and C disjoint? _yes_ Why or why not? _Even die rolls do not overlap with 5.__ |
(e) |
Are A and C independent? _no__ Why or why not? _P(A | C) = 0 ¹ P(A) = 1/2_ |
(f) |
Compute the probability of getting an even die roll and two heads, or of getting a die roll that is not a 5. Answer: 5/6 (construct a Venn diagram, or ask in class) |
(g) |
Compute the probability of getting a 5, given that the roll was not even or the coins both came up heads. Answer: 4/15 (be sure to ask in class if you can’t get this). |
2. |
The fiendish Mr. Han Sen, a lecturer in statistics at Hong Kong University, was fond of conducting quizzes to discourage students from skipping (i.e., ditching) required assemblies. For examples, after a lecture on ethics, Han Sen would often ask the students to summarize the key points of the lecture. Needless to say, students hated these quizzes and conspired to foil the evil Han Sen’s tactics. |
(a) |
You may assume for this problem that HKU has a huge enrollment (many thousands of students). Make a tree diagram. (Done in class 12/6/00). Then find or compute each of the following: |
(b) |
the sensitivity of the quiz = P(fail | cut) = 0.70 |
(c) |
the specificity of the quiz = P(pass | attend) = 0.99 |
(d) |
the probability that a student is accused by Han Sen of skipping the assembly = P(fail) = P(fail Ç attend) + P(fail Ç ~attend) = 0.0064 + 0.21 = 0.2164 |
(e) |
the probability that a student is among those who pass the quiz = P(pass) = P(pass Ç
attend) + P(pass Ç
~attend) = 0.6336 + 0.0900 = 0.7236 |
(f) |
the probability that a student is falsely accused, given that s/he actually attended the assembly = P(falsely accused | attend) = P(fail | attend) = 0.0064/0.64 = 0.01 |
(g) |
the probability that a student passed the quiz or avoided taking the quiz altogether = P(pass È
excused) = P(pass) + P(excused) = 0.7236 + 0.06 = 0.7836 |
(h) |
the probability that a student skipped the assembly, given that the student did not have an excused absence = P(skip | ~excused) = P(skip Ç ~excused) / P(~excused) = 0.3/0.94 = 0.3191 |
(i) |
the probability that a student actually skipped the assembly, given that s/he failed the quiz = P(skip | fail) = P(skip Ç fail) / P(fail) = P(skip) · P(fail | skip) / P(fail) = (0.3)(0.7)/0.2164 = 0.9704 |