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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

    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

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    स्रोत : socratic.org

    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?

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    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
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