This problem is solved using a Poisson distribution. This is because we have an average number of events happening in a certain time frame. In this case, that time frame is 1 day.
In this case, the average is mu = 2, which is the number of tools rented per day.
What we need to do is find P(x = 0), P(x = 1) and P(x = 2) which represent the probabilities of renting x = 0, x = 1, and x = 2 tools out respectively. Once we know those three items, we add them up to get P(x <= 2). This is the probability of getting at most 2 requests.
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Evaluate the probability when k = 0
P(x = k) = ( mu^k*e^(-mu) )/(k!)
P(x = k) = ( 2^k*e^(-2) )/(k!)
P(x = 0) = ( 2^0*e^(-2) )/(0!)
P(x = 0) = 0.135335
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Repeat for k = 1
P(x = k) = ( 2^k*e^(-2) )/(k!)
P(x = 1) = ( 2^1*e^(-2) )/(1!)
P(x = 1) = 0.270671
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Repeat for k = 2
P(x = k) = ( 2^k*e^(-2) )/(k!)
P(x = 2) = ( 2^2*e^(-2) )/(2!)
P(x = 2) = 0.270671
We get the same result because (2^1)/(1!) = (2^2)/(2!)
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Now add up those three results
P(x <= 2) = P(x = 0) + P(x = 1) + P(x = 2)
P(x <= 2) = 0.135335 + 0.270671 + 0.270671
P(x <= 2) = 0.676677
The probability that the firm can meet demand is roughly 0.676677
Round this however you need to
