Definitions 1
What is the range of probability?
0-1
Flip two coins and find the probabilities of the events:
Let G = the event of getting two faces (head-head, tail-tail) that are the same.
Answer: P(G) = 2/4
A box has two balls, one white, and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
c. Let H = the event of getting white on the first pick.
Answer: P(H) = ½
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
Answer: P (A or B) = P(A) + P(B) – P(A and B) = 0.65 + 0.65 – 0.585 = 0.715
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
Answer: P(B) = 0.143; P(N) = 0.85
Measures the likelihood that an event will occur
Probability
Flip two coins and find the probabilities of the events:
Answer: ½ * 1 = ½
Note: 1 is independent from the first flip
A box has two balls, one white, and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
a. Let F = the event of getting the white ball twice.
Answer: ½ * ½ = ¼
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska.
Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35. What is the probability that he does not choose to go anywhere for his vacation?
Answer: P (A or B) = P(A) + P(B) = 0.6 + 0.35 = 0.95
1 – 0.95 = 0.05 not choosing anywhere for his vacation
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
Answer: P(N|B) = 0.02
Probability that an event will occur given that another event has already occurred
Conditional Probability
Flip two coins and find the probabilities of the events:
Answer: Getting all tails occurs when tails show up on both coins (TT). H’s outcomes are HH and HT. J and H have nothing in common, so P(J and H) = 0. J and H are mutually exclusive.
Let event C = taking an English class. Let event D = taking a speech class.
Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225.
Justify your answers to the following questions numerically.
Answer: Yes, because P(C|D) = P(C)
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
Answer: P(A and B) = P (B and A). Since P (B|A) = 0.90: P (B and A) = P (B|A) P(A) = 0.90*0.65 = 0.585
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
Answer: P(B AND N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029
Probability theory is the basis for ________.
Inferential Statistics
A box has two balls, one white, and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
b. Let G = the event of getting two balls of different colors.
Answer: possible outcome – white-red, red-white (same probability)
P(white-red) = P (white on first draw) x (red on second draw) = ½ * ½ = ¼
Since both outcomes have same possibility
P(G) = 2 (P(white-red)) = 2 * ¼ = ½
Let event C = taking an English class. Let event D = taking a speech class.
Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225.
Justify your answers to the following questions numerically.
Answer: No, because P(C and D) is not equal to zero.
probability of both events occurring is not possible
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
Answer: No, they are not because P(A and B) = 0.585
To be mutually exclusive, P (A and B) must be equal to zero.
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
Answer: P(B OR N) = P(B) + P(N) - P(B AND N) = 0.143 + 0.85 - 0.0029 = 0.9901
Define independent events& dependent events
Independent: occurrence of one event does not change the probability of the occurrence of the other event
Dependent: occurrence of one event affects the probability of the occurrence of another event
A box has two balls, one white, and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
d. Are F (event of getting the white ball twice) and G (event of getting two balls of different colors) mutually exclusive?
Answer: Events F and G are not mutually exclusive because its possible to get two ball of same color (Event G) and also getting the white ball twice (Event F)
A box has two balls, one white, and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
e. Are G (event of getting two balls of different colors) and H (event of getting white on the first pick) mutually exclusive?
Answer: Events G and H are not mutually exclusive. Possible to get 2 balls of different colors (Event G) while also getting white on the first pick (Event H)
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.
Answer: No, they are not, because P(B and A) = 0.585
P(B) P(A) = 0.65*0.65 = 0.423
0.423 not equal to 0.585 = P (B and A)
So, P (B and A) is not equal to P(B) P(A)
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
Answer: No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N).