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# the probability that a contractor will get a plumbing contract is 2/3 and the probability that he will not get an electric contract is 5/9. if the probability of getting any one contract is 4/5, what is the probability that he will get both the contract?

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## The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both?

19/45, The procedure is explained below. A is the event of getting a plumbing contract and p(A) = 2/3. B is the event of getting electrical contract and p(B) = 5/9. AuuB is the event of getting atleast one of them p(AuuB) = 4/5. By addition theorem of probability, P(AuuB) = p(A) + p(B) - p(AnnB). where p(AnnB) is the probability of getting both the contracts. P(AnnB) = p(A) + p(B) - P(AuuB). i.e. p(AnnB) =2/3 + 5/9 - 4/5 or (30 + 25 - 36)/45 p(AnnB) = 19/45

## The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both?

Statistics

Ramchandran A. Nov 3, 2015 19 45

, The procedure is explained below.

### Explanation:

A is the event of getting a plumbing contract and p(A) =

2 3 .

B is the event of getting electrical contract and p(B) =

5 9 . A ∪

B is the event of getting atleast one of them

p(A ∪ B) = 4 5 .

By addition theorem of probability,

P(A ∪

B) = p(A) + p(B) - p(A

∩ B). where p(A ∩

B) is the probability of getting both the contracts.

P(A ∩

B) = p(A) + p(B) - P(A

∪ B). i.e. p(A ∩ B) = 2 3 + 5 9 - 4 5 or 30 + 25 − 36 45 p(A ∩ B) = 19 45 Answer link

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## The Probability that a Contractor Will Get a Plumbing Contractor is 2 3 . and the Probability that He Will Not Get an Electric Contract is 5 9 . If the Probability of Getting at Least One Contrac

The Probability that a Contractor Will Get a Plumbing Contractor is 2 3 . and the Probability that He Will Not Get an Electric Contract is 5 9 . If the Probability of Getting at Least One Contrac

The probability that a contractor will get a plumbing contractor is

23

. and the probability that he will not get an electric contract is

59

. If the probability of getting at least one contract is

45

what is the probability that he will get both?

### SOLUTION

1445

Explanation:

Let the probability of getting plumbing contract P(A) =

23

and the probability of getting an electric contract P(B) = 1 - P (not getting electric contract)

= 1-59=49 Also, P(A∪B) = 45

∴ Required probability, P (A∩B)

= P(A) + P(B) - P(A∪B)

= 23+49-45 = 30+20-3645=1445

Concept: Probability (Entrance Exam)

Is there an error in this question or solution?

स्रोत : www.shaalaa.com

## The probability of a contractor getting a plumbing contract is 2/3 and the probability of him getting an electricity contract is 5/9. The probability of getting at least one contract is 4/5. What’s the probability that he he will get both contracts?

Answer (1 of 4): The probability of getting at least one contract [P(AUB)] is equal to the probability of getting an electrical contract [P(A)] plus the probability of getting the plumbing contract [P(B)] minus the probability of getting both contracts [P(A∩B)]. In math symbols: P(AUB) = P(A) + ...

The probability of a contractor getting a plumbing contract is 2/3 and the probability of him getting an electricity contract is 5/9. The probability of getting at least one contract is 4/5. What’s the probability that he he will get both contracts?

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Sort Steven Smith

Earned 98% or higher in all my math classes at UCMO.Author has 3.3K answers and 6.6M answer views5y

Defining A A

as the contractor getting a plumbing contract, and

B B

as the contractor getting an electricity contract, we have:

P(A)= 2 3 P(A)=23 P(B)= 5 9 P(B)=59 P(A∪B)= 4 5 P(A∪B)=45

For any 2 events, whether or not they are independent, this identity holds true:

P(A∪B)=P(A)+P(B)−P(A∩B)

P(A∪B)=P(A)+P(B)−P(A∩B)

Plugging in the values that we know from the problem:

4 5 = 2 3 + 5 9 −P(A∩B) 45=23+59−P(A∩B) P(A∩B)= 2 3 + 5 9 − 4 5 P(A∩B)=23+59−45 P(A∩B)= 30 45 + 25 45 − 36 45

P(A∩B)=3045+2545−3645

P(A∩B)= 30+25−36 45 = 19 45

P(A∩B)=30+25−3645=1945

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Let P P

be the event that the contractor gets the plumbing contract,

E E

be the event that he gets the electricity contract.

We are given Pr(P)= 2 3 Pr(P)=23 , Pr(E)= 5 9 Pr(E)=59 and Pr(P∪E)= 4 5 Pr(P∪E)=45 . To find Pr(P∩E) Pr(P∩E)

, we will use the following equality

Pr(P∩E)=Pr(P)+Pr(E)−Pr(P∪E)=

2 3 + 5 9 − 4 5 = 19 45

Pr(P∩E)=Pr(P)+Pr(E)−Pr(P∪E)=23+59−45=1945

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Robert Nichols

Author has 5K answers and 11.2M answer views4y

The probability of getting at least one contract [P(AUB)] is equal to the probability of getting an electrical contract [P(A)] plus the probability of getting the plumbing contract [P(B)] minus the probability of getting both contracts [P(A∩B)].

In math symbols: P(AUB) = P(A) + P(B) - P(A∩B). So we can isolate P(A∩B). Add P(A∩B) to both sides and subtract P(AUB) from both sides.

P(A∩B) = P(A) + P(B) - P(AUB)

P(A∩B) = 5/9 + 2/3 - 4/5 convert to a common denominator of 45

P(A∩B) = 25/45 + 30/45 - 36/45 = 19/45

Stephen Rumberg

QuickBooks Statement Converter Developer (2011–present)Author has 2.2K answers and 1.4M answer views5y

Zero. The poor guy is going to spend so much time figuring out probability that he’ll forget to bid the jobs on time!

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Related

The probability that a student passes mathematics is 2/3 and the probability that he passes English is 4/9. If the probability of passing at least one course is 4/5, then what is the probability that he will pass both mathematics and English?

P(passes M) = 2/3 P(passes E) = 4/9

P( passes M U passes E) = P(M) + P(E) - P(M and E)

4/5 = 2/3 + 4/9 - P( passes M and E )

P(passes M and E ) = 2/3 + 4/9 - 4/5 = 30/45 + 20/45 - 36/45

used 45 as common denominator

स्रोत : www.quora.com

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Mohammed 1 month ago

Guys, does anyone know the answer?