What is. 9r-5r+2+r? ??????????

If a watch has a 0.12 defect rate. what is the probability that the watch has fewer than 3 defects in a run of 75 chimes?

Maslowich

Answer:

$(\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

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 $(\frac{88}{100})^{75}$ , because for every chime there is a probability of $\frac{88}{100}$ that there is no defect

2.) The probability that the watch has exactly 1 defect is $(\frac{88}{100})^{74}(\frac{12}{100})$ times the number of ways you can choose 1 of 75 of the chimes to be defective, which is $\binom{75}{1}$ , so the probability that the watch has exactly 1 defect is $\binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100})$ .

3.) For the same reason as 2.), the probability that the watch has exactly two defects is $\binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

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 $(\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}$

8 0

4 years ago

10 points plz answer thx

amid [387]

Answer:

$( \sqrt[4]{81} ) {}^{3}$