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A computer consulting firm presently has bids out on three projects. Let $A_{i}=\{$ awarded project $i\}$

for $i=1,2,3,$ and suppose that $P\left(A_{1}\right)=.22, P\left(A_{2}\right)=.25, P\left(A_{3}\right)=.28, P\left(A_{1} \cap A_{2}\right)=.11$ $P\left(A_{1} \cap A_{3}\right)=.05, P\left(A_{2} \cap A_{3}\right)=.07, P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 .$ Express in words each of the following events, and compute the probability of each event:

(a) $A_{1} \cup A_{2}$

(b) $A_{1}^{\prime} \cap A_{2}^{\prime}$

(c) $A_{1} \cup A_{2} \cup A_{3}$

(d) $A_{1}^{\prime} \cap A_{2} \cup A_{3}$

(e) $A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}$

(f) $\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}$

Answer: P(A1 ? A2) = 0.35

Solution:

A1 : '' awarded project 1''

A2 : ''awarded project 2''

A3: ''awarded project 3''

In set theory we write the union of events A and B as A?B.

A?B means that the event A occurs,event B occurs or either both events occurs at the same time.

The probability is given by the equation :

P(A?B) = P(A) + P(B) - P(A?B) (1)

Where the event (A?B) is the event where A and B occur at the same time

and P(A?B) is the probability of (A?B)

Using the equation (1) :

P(A1 ? A2) = P(A1) + P(A2) - P(A1?A2)

P(A1 ? A2) = 0.22 + 0.25 - 0.12

P(A1 ? A2) = 0.35

Probability Topics

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Md. 1.

April 17, 2021

find the conditional probability and interpret the probability

in this question. We have a firm that is bidding on three projects, and we have event. A one is that they are awarded the first project. A two is there, awarded the second project, and a three is there awarded third project. And so we have the probabilities for these three events. And we also have the probabilities for the intersections of these events, for example, being awarded the probability of being awarded both the first and the second contract, or being awarded the 1st and 3rd contract, etcetera and then were asked, were given some combinations of events and asked you to describe these events qualitatively and to also calculate the probabilities of these events. So for part A, we have a one union with a two, so we can describe this as the firm being awarded at least one of the first two projects, and the other part is to calculate the probability of this event. And so most of these questions use either the the addition Law or the Compliment Law for probability. So this one makes use of the addition law, which says that this probability is equal to the probabilities of the individual events summed, minus the probability of the intersection of these individual events. So we have. We have these three probabilities in the right side of the equation given to us in the question. We have 0.22 plus 0.25 minus 0.11 gives us a probability of 0.36 So here we have a compliment intersected with this is a one compliment intersected with a to complement this event is the firm being awarded neither of the first two contracts. And to calculate this probability, we can make use of Dem Organs Law, which tells us that tells us that this is equal to the probability of a one union A to all complimented. And so the conflict that we're looking at, the comp, the probability of the compliment of a one union union with a two. So we've already found this probability of a one union with a two in part A, which is 0.36 compliment is one minus that probability, which gives us 0.64 and I should probably have an extra set of brackets here. So it's clear that the compliment applies to everything inside the inner set of brackets. The event for Part C is a one union with a to union with a three. This can be described as thief ERM being awarded at least one of the three projects, and it's probably probability can be solved using the addition rule again, but this time for the union of three events. So we had the probabilities of the individual events. Then we subtract the probabilities of each of the intersection of each combination of two events. So we subtract the probability of way a one intersected with a two and subtract the probability of a one intersected with a three and subtracted probability of a two intersected with a three. And then we have to add back. The probability of all three events intersected. And if these formulas don't make sense to you, I suggest that you look at the Venn diagrams that air presented in the textbook. For this chapter, it helps to illustrate exactly why we're subtracting and then adding these intersections of different events. So if we had the numbers that are given in the question for these probabilities, we get the following. This comes out to a probability of 0.53 for party. The event is a one compliment intersected with a to complement intersected with a three compliment, and this could be described as the firm being awarded none of the contracts. This one is a little easier to think of. Qualitatively have the the firm being awarded. None of the contracts is the complement of the firm being awarded at least one of the contracts. And so we can say that this probability is equal to one, minus the probability that it is awarded at least one of the contracts which we have already solved for. So it is one minus 0.53 That gives us 0.47 He is a one compliment intersected with a to complement, intersected with a three, and this event can be described qualitatively as being awarded contract number three. But neither of the first two contracts. Now it helps a lot to use a Venn diagram when solving more complicated events such as this one. So in this diagram, the region of interest is a three but not a one and a two. So that is this region, so that could be thought of as the probability of a three minus the probability of the intersection of a one and a three. So we take all of the area and a three, and then we subtract the intersection of way a one and a three, and then we're going to subtract the intersection of a two and a three. But then, because we have subtracted this area in here twice, we have to add it back. So we add that last section back with this term, that section is the intersection of a one, a two and a three, and we have all of these probabilities from the question. This comes out to 0.17 for if the event is a one, Compliment intersected with a to complement in brackets Union with a three. And this event can be described in words as awarded neither of contracts one and two or awarded contract number three. So here, neither a one nor a to is everything outside of a one and a two. It's everything that is outside of both a one and a two. So that is this region here, and then if we union that with a three, So that's the last part or a three. It means everything in red as well as whatever is in pink. So pink is all of a three. So our region of interest is everything except for the shaded areas. So the region of interest is this area here inside the outline. And so the probability of this event could be thought of the as the probability that none are awarded, plus the probability that the third event is awarded. We already have this probability from Part D, and we have the probability of being awarded the third contract from the question 0.28 This comes out to 0.75