The probability that a contractor will get a | vedclass

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(C) Let $A$ be the event that the contractor gets a plumbing contract and $B$ be the event that he gets an electric contract.Given: $P(A) = 2/3$.The p

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$14/45$

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(C) Let $A$ be the event that the contractor gets a plumbing contract and $B$ be the event that he gets an electric contract.Given: $P(A) = 2/3$.The probability that he will not get an electric contract is $P(\bar{B}) = 5/9$. Therefore,$P(B) = 1 - P(\bar{B}) = 1 - 5/9 = 4/9$.The probability of getting at least $1$ contract is $P(A \cup B) = 4/5$.We use the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $4/5 = 2/3 + 4/9 - P(A \cap B)$.$P(A \cap B) = 2/3 + 4/9 - 4/5$.Finding a common denominator $(45)$: $P(A \cap B) = (30 + 20 - 36) / 45 = 14/45$.Thus,the probability that he will get both contracts is $14/45$.

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 at least $1$ contract is $4/5$,what is the probability that he will get both?

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