Question

Suppose that we have a fuse box containing 40 fuses of which 6 are defective. If two fuses are selected at random and removed from the box. Find the probability that they both are defective, if the first fuse Replaced Not replaced

Answer 1

Answer:

For this question I know that one must use the formula

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=5

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A) and P(B|A)=

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A) and P(B|A)=P(A∩B)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A) and P(B|A)=P(A∩B)p(A)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A) and P(B|A)=P(A∩B)p(A)

For this question I know that one must use the formulaP(A∩B∩C)=P(A)∗P(B|A)∗P(C|A∩B)I am able to to identify that P(A)=520 . I assume since their are five out of 20 defective fuses. My question for this probelm is how does one identify P(B|A) and P(C|A∩B)?I know that P(B|A)=P(B∩A)p(A) and P(B|A)=P(A∩B)p(A) Unfortunately identifying how one solves this problem eludes my understanding any advice on how to solve this would be beneficial to I.

Explanation:

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