# 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?

### Mohammed

Guys, does anyone know the answer?

get 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? from screen.

## 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

### 1 Answer

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

### Related questions

How do I determine the molecular shape of a molecule?

What is the lewis structure for co2?

What is the lewis structure for hcn?

How is vsepr used to classify molecules?

What are the units used for the ideal gas law?

How does Charle's law relate to breathing?

What is the ideal gas law constant?

How do you calculate the ideal gas law constant?

How do you find density in the ideal gas law?

Does ideal gas law apply to liquids?

### Impact of this question

45777 views around the world

You can reuse this answer

Creative Commons License

## 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

Advertisement Remove all ads MCQ

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?

### OPTIONS

1645 1345 1445 1572 Advertisement Remove all ads

### 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?

Advertisement Remove all ads Advertisement Remove all ads

## 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?

Ad by Adclickersbot

Make Money Easy And Safe Way.

Adclickersbot is a Telegram earning platform that allows you to earn money by performing simple tasks.

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

Related questions

The probability that a contractor will get a contract A is 2/3 and contract B is 3/5, then what is the probability that he will get both contracts and the probability that he will not get any contract?

Probability that a contractor will get a contract for road construction is4/9 and the probability that he will get contacts for the construction of water tank is 5/7. What is the probability of getting at least one of the contract.?

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?

A company tenders for two contract A and B. The probability that it will obtain A is 0.2 and contract B is 0.3. what will is the probability that it will obtain either contract A or contract B?

There are 2 boxes with 1 containing 4 red and 3 green balls and the other containing 3 blue and 4 green balls. One ball is drawn at random from each box. What is the probability of 1 ball red and other ball blue?

Amit Goyal

Founder and Instructor at Econ School (2009–present)Author has 644 answers and 3.1M answer views5y

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

Sponsored by Grammarly

Free English writing tool.

Write in clear, mistake-free English with our free writing app. Try now!

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!

Ags Ilango 4y Related

In a class there are 15 boys and 10 girls. Three students are selected at random. What is the probability that one girl and 2 boys are selected?

Sponsored by USAFIS

This is the best time to apply for the Green Card DV Lottery!

Get a chance to win and apply today! America is waiting for you with many amazing opportunities.

Alan Silver

Studied Electrical Engineering (Graduated 1984)Author has 1.1K answers and 676.8K answer views3y

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

Guys, does anyone know the answer?