Question 10.4: The probability that a printer will need correcting adjustme......
The probability that a printer will need correcting adjustments during a day is 0.2. If there are five printers running on a particular day determine the probability of:
1. No printers need correcting
2. One printer needs correcting
3. Two printers require correcting
4. More than two printers require adjusting.
There are five trials (with n = 5, p = 0.2, and the success of a trial is a printer needing adjustments). And so:
1. This is given by P(X=0)={\binom{5}{0}}\ 0.2^{0}\ast0.8^{5}=0.3277
2. This is given by P(X=1)={\binom{5}{1}}\ 0.2^{1}\ast0.8^{4}=0.4096
3. This is given by P(X=2)={\binom{5}{2}}\ 0.2^{2}\ast0.8^{3}=0.205
4. This is given by 1 − P(2 or fewer printers need correcting)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 – [0.3277 + 0.4096 + 0.205]
= 1 – 0.9423
= 0.0577