There are three workstations available having steady-state probabilities of 0.99, 0.95, 0.85 of being available on demand. What is the probability that at least two of the three will be available at any given time?

99.065% probability that at least two of the three will be available at any given time.

Step-by-step explanation:

We have these following probabilities:

99% probability of the first workstation being available

95% probability of the second workstation being available

85% probability of the third workstation being avaiable

Two being available:

We can have three outcomes

First and second available, third not. So

0.99*0.95*0.15 = 0.141075

First and third available, second not. So

0.99*0.05*0.85 = 0.042075

Second and third available, first not. So

0.01*0.95*0.85 = 0.008075

Adding them all

P(2) = 0.141075 + 0.042075 + 0.008075 = 0.191225

Three being available:

P(3) = 0.99*0.95*0.85 = 0.799425

What is the probability that at least two of the three will be available at any given time?

P = P(2) + P(3) = 0.191225 + 0.799425 = 0.99065

99.065% probability that at least two of the three will be available at any given time.

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7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25