Answer:
[tex](\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}[/tex]
Step-by-step explanation:
If a watch has fewer than three defects, then either 1.) It has no defects, 2.) it has exactly 1 defect, or 3.) it has exactly 2 defects.
1.) The probability that the watch has no defects is [tex](\frac{88}{100})^{75}[/tex], because for every chime there is a probability of [tex]\frac{88}{100}[/tex] that there is no defect
2.) The probability that the watch has exactly 1 defect is [tex](\frac{88}{100})^{74}(\frac{12}{100})[/tex] times the number of ways you can choose 1 of 75 of the chimes to be defective, which is [tex]\binom{75}{1}[/tex], so the probability that the watch has exactly 1 defect is [tex]\binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100})[/tex].
3.) For the same reason as 2.), the probability that the watch has exactly two defects is [tex]\binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}[/tex]
Since 1.), 2.), and 3.) are mutually exclusive events, the probability of their union is simply the probability of each of them added together, which is [tex](\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}[/tex]
