Saying that a set of events is mutually exclusive and collectively exhaustive implies that one and only one of the events can occur on any trial.
True
The sum of the probabilities of the occurrence of the event in the case of mutually exclusive and collectively exhaustive is 1. When the set of events is mutually exclusive, both cannot occur at the same time while when the set is collectively exhaustive, at least one of the events must occur. Hence combining the two, the above statement holds true.
Saying that a set of events is mutually exclusive and collectively exhaustive implies that one and...
IS THIS CORRECT? What is the difference between mutually exclusive events and collectively exhaustive events? Choose the correct choice below. O A. A set of events are mutually exclusive if the probability of each event in the set is not affected by the outcomes of the other events. A set of events are collectively exhaustive if at least one of the events must occur. OB. A set of events are mutually exclusive if they cannot occur at the same time....
Come up with an example of two mutually exclusive and collectively exhaustive events of your choice and assign their probabilities. Come up with a conditioning event and demonstrate how conditioning can change the probability of these events. Explain why that is the case. 2.
Question 13 (2 points) If two events are mutually exclusive, then their probabilities can be added all of the choices are correct they may also be collectively exhaustive if one occurs, the other cannot occur the joint probability is equal to O Question 14 (1 point) The area under the normal distribution to the left of the mean is always 0.5 or 50 percent. True False
1) Events A, B, C, and D are mutually exclusive and collectively exhaustive. If P(A or B) = .44. Circle all of the following that are possible? a) P(C or D) > .56 b) P(C and D) = .56 c) P(A and D) = 0 d) P(A) = .5 2) You have 10 members of a team. How many ways could you pick 3 captains (order does not matter)? a) 720 b) 30 c) 120 ...
All electric circuit board failures can be classified into one of the following mutually exclusive and collectively exhaustive categories: assembly defects (40%), electrical component defects (35%), and mechanical defects (25%). Suppose ten boards fail independently, and let the random variables X, Y, and Z denote the number of each type of defect among the boards respectively. What is P{X=8,Z=1|Y=1}?
Let a sample space be partitioned into three mutually exclusive and exhaustive events, py, 2, and,Ie. Complete the following probability table. (Round your answers to 2 decimal places.) Conditional Probabilities Prior Joint P(Bi) 0.11 PIA B)0.46 P(An B) P(B I A) P(83 1 A) Total P(A)
11)IfA, B and Care mutually exclusive and exhaustive events andP(A)= P (B)= P(C, find P (A), P (B) and P (C)
From the following events, judge whether the pair of events are: (i) mutually exclusive or not mutually exclusive and (ii) collective exhaustive or not collective exhaustive. 1. Drawing a king or an ace from a deck of cards. 2. Getting a yellow ball or a red ball from an urn of yellow and red balls. 3. Activity A or Activity B that start simultaneously in a given project containing 10 activities. 4. Activity A or Activity D that start at...
Let a sample space be partitioned into three mutually exclusive and exhaustive events, B1, B2, and, B3. Complete the following probability table. (Round your answers to 2 decimal places.) Prior Probabilities Conditional Probabilities Joint Probabilities Posterior Probabilities P(B1) = 0.11 P(A | B1) = 0.45 P(A ∩ B1) = P(B1 | A) = P(B2) = P(A | B2) = 0.62 P(A ∩ B2) = P(B2 |A) = P(B3) = 0.38 P(A | B3) = 0.85 P(A ∩ B3) = P(B3...
Let a sample space be partitioned into three mutually exclusive and exhaustive events, B1, B2, and, B3. Complete the following probability table. (Round your answers to 2 decimal places.) Prior Probabilities Conditional Probabilities Joint Probabilities Posterior Probabilities P(B1) = 0.15 P(A | B1) = 0.40 P(A ∩ B1) = P(B1 | A) = P(B2) = P(A | B2) = 0.65 P(A ∩ B2) = P(B2 |A) = P(B3) = 0.32 P(A | B3) = 0.75 P(A ∩ B3) = P(B3...