2. A machine shop has 100 drill presses and other machines in constant use. The

Question

2. A machine shop has 100 drill presses and other machines in constant use. The probability that a machine will become inoperative during a given day is 0.002. During some days, no machines are inoperative, but during some days, one, two, three, or more are broken down. What is the probability that fewer than two machines . will be inoperative during a particular day? (10) David's gasoline station offers 4 cents off per gallon if the customer pays in cash and does not use a credit card. Past evidence indicates that 40% of all customers pay in cash. During a one-hour period, 15 customers buy gasoline at this station. What is the probability that at least 10 pay in cash? 3. (10) ?#

Explanation / Answer

Solution:

2. For a Poisson distribution as described above, the probability of exactly k events during the given time period is:
P(X = k) = ( ?^k). e^- ?/k!
In this case, our time period is one day, we have 100 machines in use, and the probability of any given machine being inoperative on any given day is 0.002 (and the probabilities of any two machines being inoperative are independent). So we would expect 0.002 * 100 = 0.2 machines to be inoperative on any day (obviously, 0.2 machines cannot be inoperative --- think 2 machines inoperative every 10 days). So ? = 0.2.
The probability of fewer than two machines being inoperative is equal to the probability of zero machines being inoperative plus the probability of one machine being inoperative.
P(X < 2) = P(X = 0) + P(X = 1)

Applying our formula, we get:
P(X < 2) = ( ?^0). e^- ?/0! + ( ?^1). e^- ?/1!
Substituting in for ? and simplifying, we get:
( 0.2^0). e^-0.2/0! + ( 0.2^1). e^-0.2/1! (simplifying 0.2 for ?)
e^-0.2+ 0.2. e^-0.2 (simplifying 0! = 1, 1! = 1, and the exponents)
1.2. e^-0.2 (combining like terms)
So the probability of fewer than two machines being inoperative on any given day is 1.2. e^-0.2 or (to the nearest thousandth) 0.982. This makes sense: if we only expect 2 machines to be inoperative every ten days, the chances of two machines being inoperative in one day should be very small.

3. Note that P(at least x) = 1 - P(at most x - 1).   
  
Using a cumulative binomial distribution table or technology, matching   
  
n = number of trials = 15   
p = the probability of a success = 0.4   
x = our critical value of successes = 10   
  
Then the cumulative probability of P(at most x - 1) from a table/technology is   
  
P(at most 9 ) = 0.966166697
  
Thus, the probability of at least 10 successes is  
  
P(at least 10 ) = 0.033833303

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