Suppose you just received a shipment of thirteen TV's. Two of the TV's are defective. If two TV's are randomly selected, compute the probability that both TV's work. What is the probability at least one does not work? formula: P(E) * P(F|E)
(answers are in fractions)
Works: 55/78
Not works: 23/78
I'm not sure how they got these answers. Can someone show me the steps in how they got these? Thanks!
(answers are in fractions)
Works: 55/78
Not works: 23/78
I'm not sure how they got these answers. Can someone show me the steps in how they got these? Thanks!
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No. of TVs work = 13-2 = 11
No. of TVs not work = 2
i) P(E) = P(selecting the first TV that works) = 11/13
P(F/E) = P(selecting the second TV that works) = (11-1)/(13-1) = 10/12
According to the Multiplication Theorem of Probability
P(selecting two TVs that work) = P(E)*P(F/E) = 11/13 * 10/12 = 110/156 = 55/78
ii) P(selecting two TVs such that at least one TV does not work) = 1 - P(selecting two TVs that work)
= 1 - 55/78
= 23/78
Remember that the Sample space for the Number of TVs that work is (0 1 2)
Further the sum of the probabilities of all possible outcomes of a random variable = 1
i.e., P(0)+P(1)+P(2) = 1
Therefore P(0)+P(1) = 1 - P(2)
No. of TVs not work = 2
i) P(E) = P(selecting the first TV that works) = 11/13
P(F/E) = P(selecting the second TV that works) = (11-1)/(13-1) = 10/12
According to the Multiplication Theorem of Probability
P(selecting two TVs that work) = P(E)*P(F/E) = 11/13 * 10/12 = 110/156 = 55/78
ii) P(selecting two TVs such that at least one TV does not work) = 1 - P(selecting two TVs that work)
= 1 - 55/78
= 23/78
Remember that the Sample space for the Number of TVs that work is (0 1 2)
Further the sum of the probabilities of all possible outcomes of a random variable = 1
i.e., P(0)+P(1)+P(2) = 1
Therefore P(0)+P(1) = 1 - P(2)