Question #17692

suppose you receive a shipment of ten televisions. Three of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work?

Expert's answer

P(both work) = ((10-3)/10)*((10-3-1)/9) = (7/10)*(6/9) = 42/90

P(at least one does not work) = 1 - P(both work) = 1 - 42/90 = 48/90 = 24/45.

P(at least one does not work) = 1 - P(both work) = 1 - 42/90 = 48/90 = 24/45.

## Comments

Assignment Expert18.10.19, 15:47The probability that at least one of the two televisions does not work is 24/45.

davey18.10.19, 04:21Suppose you just received a shipment of tenten televisions. ThreeThree of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work The probability that at least one of the two televisions does not work is

Assignment Expert15.10.19, 17:09Dear Tikki. Please use the panel for submitting new questions. The required probability is P(1 woks or 2 works)=P(1 works)+P(2 works)- P(1 works and 2 works)=(1-0.18)+(1-0.18)-(1-0.18)(1-0.18)=0.9676.

Tikki15.10.19, 09:08For a parallel structure of identical components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has 0.18 probability of failure. What is the probability that a parallel structure with 2 identical components will succeed?

Assignment Expert28.02.19, 15:24Dear Mariana,

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Mariana28.02.19, 01:19The way he wrote out how he solved this question really helped me do questions on my homework. Therefore i would like to thank the expert's answer for this question.

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