C784 MODULE 7.11 Conditional probability
4. What are dependent events?
Dependent events are not independent by definition. The probability of a dependent event is also dependent on another event. Therefore, all of the above is correct.
GENERAL MULTIPLICATION RULE dependent events
is the probability of A and B = the probability of A times the probability of B given A. P(A and B) = P(A) x P(B[A) A= B= P(A)= P(B)= P(A and B)= P(B[A)=
1) What do we call two events for which the occurrence of the first affects the probability that the second event occurs?
dependent events If the occurrence of the first affects the probability that the second event, they are known as dependent events
4. Two events can be independent and dependent. True or False?
false Two events can be either independent or dependent.
2) You select one card from a deck of cards, and do NOT place that card back in the deck. Then, you select a second card from the deck of cards. Determine if the following two events are independent or dependent: Selecting a queen and then selecting a king.
DEPENDENT The occurrence of the first affects the probability that the second event, therefore they are known as dependent events.
DEPENDENT EVENTS vs INDEPENDENT EVENTS
DEPENDENT *Flipping a heads on the first coin toss and flipping 2 heads in a row *Living in Africa and having dark skin *Loss of a parent or loved one as a child and clinical depression as an adult. *Not purchasing any raffle tickets and winning the raffle. INDEPENDENT *Being born in Minnesota and being a girl. *Rolling a 6 on the first roll and rolling a 6 on the second roll *Being shorter than average and having at least 10 close friends.
5. Independence can be determined from any of the four given statements below: P(A and B) = P(A) · P(B) P(A | B) = P(A) P(B | A) = P(B) P(B | A) = P(B | not A) True or False?
. True Correct. This is a true statement. Any of the four cases determine if the two events are independent.
4) A and B are independent events. P(A) = 0 and P(B) = 1. Calculate P(A and B)
0 P(A and B) = P(A) · P(B) = 0 · 1 = 0.
5) A and B are independent events. P(A) = 0 and P(B) = 0.2. Calculate P(A | B)
0 P(A | B) = P(A) = 0
6) A and B are independent events. P(A) = 0.3 and P(B) = 0. Calculate P(B | A).
0 P(B | A) = P(B) = 0.
4) A and B are independent events. P(A) = 1 and P(B) = 0.1. Calculate P(A and B).
0.1 P(A and B) = P(A) · P(B) = 1 · 0.1 = 0.1
6) A and B are independent events. P(A) = 0.8 and P(B) = 0.2. Calculate P(B | A).
0.2 P(B | A) = P(B) = 0.2.
5) A and B are independent events. P(A) = 0.3 and P(B) = 0.8. Calculate P(A | B).
0.3 P(A | B) = P(A) = 0.3.
5) A and B are independent events. P(A) = 0.7 and P(B) = 0.5. Calculate P(A | B).
0.7 P(A | B) = P(A) = 0.7.
4) A and B are independent events. P(A) = 1 and P(B) = 0.9. Calculate P(A and B).
0.9 P(A and B) = P(A) · P(B) = 1 · 0.9 = 0.9.
A and B are independent events. P(A) = 0.5 and P(B) = 1. Calculate P(B | A).
1 P(B | A) = P(B) = 1
If there are 6 red scrubs and 7 green scrubs, what is the probability that a red scrub is chosen and then another red scrub?
5/26 The probability that one red scrub is chosen is 6/13 The probability that a second red scrub is chosen is 5/12 Canceling the 6 and reducing the 12 leaves a reduced fraction of 5/26 P(A & B) = P(A) x P(B)
2. There is a jar with 11 marbles: seven red marbles and four green ones. Your sister then takes a red marble and still has it. What is now the probability of choosing a green marble, if you select a marble at random?
4/10 There are now four green marbles and 10 total marbles in the jar.
in a study, 3 percent of the men studied had a systolic blood pressure of 180 mmHg, and that the probability of suffering a stroke given this high blood pressure was 18 percent. What's the probability of both having very high blood pressure and suffering stroke?
A= having a systolic # of 180 mmHg or above B= suffering a stroke P(A)= the probability of having a systolic number of 180 mmHg or above = 0.03 P(B)= is the probability of suffering a stroke P(A and B)= is the probability of having high blood pressure and suffering a stroke = ? DON'T KNOW YET P(B[A)= is the probability of suffering a stroke, given a participant has high blood pressure. = 0.18 P(A and B) = P(A) x P(B[A) P(A and B) = 0.03 x 0.18 P(A and B) = 0.0054
determining if an event is dependent or independent
Any of the following 4 formulas will establish that two events are independent. So, as usual, let A = one event B = another event. These events are independent if: P(A and B) = P(A) x P(B). P(A[B) = P(A) P(B[A) = P(B) P(B[A) = P(B[ not A). So these three last formulas should be pretty intuitive. If two events are independent then whether one has occurred will not affect the probability of the other. For example, think about it being Monday, and it's raining. If it is Monday, this isn't going to change the probability if it rains. If it rains, it doesn't change the probability that it is Monday. The P(A[B) = P(A), and the P(B[A) = P(B). So likewise, the probability of that it rains is not affected by it being some other day of the week. The P(B[A) = P(B[ not A)
4. Conditional probability cannot be used for more than two events. True or False?
FALSE Conditional probability can be used for more than two events.
Conditional Probability Rule P(B | A) = P(A and B) ---------- P(A)
Let us illustrate this rule by using a very simple example. Imagine you are flipping a coin twice. Let A = "getting a head on the first flip" and let B = "getting two heads." Let us look at a number of categories related to probability: 1. The sample space* of this event is HH, HT, TH, TT 2. P(A) is 2 possibilities out of 4 or 1/2 3. P(B) is 1 possibility out of 4 or 1/4 4. A and B, the intersection, is the set of elements that belong to both categories, which is just the element HH. So the P(A and B) = 1/4 5. If you have already flipped a head, you have 2 possibilities for the second flip, heads & tails. P(B | A) = 1/2 So substituting the probabilities above into our general conditional probability formula P(B | A) = P(A and B) ---------- P(A)
3) For independent events, what does P(A and B) equal?
P(A and B) = P(A) · P(B).
8. If you are asked to find the probability of either event A happening or event B happening, where events A and B are dependent which of the formulas below do you use?
P(A or B) = P(A) + P(B) - P(A and B)
GENERAL ADDITION RULE dependent events
P(A or B) = P(A) + P(B) - P(A and B)
3) For independent events, what does P(A | B) equal?
P(A | B) = P(A).
the probability of the intersection of two dependent events* is
P(A) × P(B | A).
the probability of the intersection of two independent events* is
P(A) × P(B)
2) For independent events, what does P(B | A) equal?
P(B | A) = P(B).
5. P(B | A) = P(A and B). True or False?
P(B | A) has a different denominator from the P(A and B) since A has already occurred.
3) For independent events, what does P(B | not A) equal?
P(B | not A) = P(B | A).
CONDITIONAL PROBABILITY RULE dependent events
P(B[A) = P(A and B)/P(A) P(B[A) -means the probability of B given A A=it rained yesterday B= I water my garden on a certain day p(b) = 0.5, water garden qoday P(B[A)= 0.1 (water garden day after it rains. don't water my garden as often when it rains) suppose we did not know the P(B[A). We record the weather for 100 days and our watering to get an experimental probability. 9/100 days it rained the day before, but we still water the garden. 24/100 days it rained. A= it rains B= water garden P(A)=prob. it rains =24/100 = 0.24 P(B)= prob. I water garden ??? don't know this yet P(A and B) = prob. it rains & water garden = 9/100 = 0.09 P(B[A) = prob. I water garden given it rained P(B[A) = P(A and B)/P(A) P(B[A) =0.09/0.24 =0.375
1. The probability of going fishing on a Saturday is represented as P(F). The probability of catching a fish on a Saturday is P(C). How would we write the probability of catching a fish on a Saturday, given that you have gone fishing?
P(C | F) This is read as "the probability of catching a fish, given that you have gone fishing."
1. What is conditional probability?
Probability of an event occurring given that another event has already occurred.
3. The intersection is always used for conditional probability. True or False?
TRUE This statement is true. Conditional probability does use the intersection.
solve for P(B | A). Write the answer as a percent rounded to the nearest tenth A=16.9% B=? P(A & B) = 4.2%
The answer is 24.9. P(B | A) = P(A and B) ---------- P(A) = 0.042 --------- = 0.2485 = 24.9% 0.169
3. The probability of the union of two dependent events is P(A) + P(B | A). True or False?
The probability is P(A) + P(B) - P(A and B).
1. The probability of two independent events both occurring is P(A) + P(B). True or False?
The probability of 2 independent events both occurring is P(A) x P(B).
2. Which statement does not belong for two independent events? a. P(A and B) = P(A) · P(B) b. P(A | B) = P(B) c. P(B | A) = P(B) d. P(B | A) = P(B | not A)
b The answer is b. The correct statement is P(A | B) = P(A).
3. If the probability of having blond hair is 5%, then the probability of having blond hair, given that you are Swedish, is 5%. True or False?
FALSE Being Swedish and having blond hair are dependent events, as Swedes are more likely to have blond hair.
5. Conditional probability does not rely on another event happening. True or False?
FALSE Conditional probability is the probability of an event occurring after one event has happened.
2. For dependent events, the probability of B is always equal to the probability of B, given A. True or False?
FALSE For dependent events, the probability of B is not always equal to the probability of B, given A.
Determining Independence
Two Events are Independent If: P(A and B) = P(A) · P(B) P(A | B) = P(A) P(B | A) = P(B) P(B | A) = P(B | not A) Please note, if any of these conditions holds, the events are independent
1) You flip a coin twice. Determine if the following two events are independent or dependent: Flipping a heads the first time and flipping a tails the second time.
independent Flipping a heads the first time does not affect the probability that you will flip a tails the second time. Therefore, they are independent events.
2) You roll a fair, six-sided die twice. Determine if the following two events are independent or dependent: Rolling a three and rolling a four
independent Rolling a three on the first roll does not affect the probability that you will roll a four on the second roll. Therefore, they are as independent events.
1) What do we call events where the occurrence of one event does not affect the probability that the other event will occur?
independent events If the occurrence of one event does not affect the probability that the other event will occur, they are known as independent events.