Question

Ests show that the first photo detector isacceptable with probability 3/5. When the first photo detector is acceptable, the second photodetector is acceptable with probability 4/5. If the first photo detector is defective, the secondphoto detector is acceptable with probability 2/5.What is the probability that exactly one photo detector of a pair is acceptable?

Answer 1

Answer:

The probability that exactly one photo detector of a pair is acceptable is $\frac{19}{25}$.

Step-by-step explanation:

Let the A = first photo detector is acceptable and B = second photo detector is acceptable .

The information provided is:

$P(A) = \frac{3}{5} \\P(B|A)=\frac{4}{5}\\ P(B|A^{c})=\frac{2}{5}$

• The law of total probability states that:

         $P(X)=P(X|Y)P(Y)+P(X|Y^{c})P(Y^{c})$

Using the law of total probability determine the probability of B as follows:

$P(B)=P(B|A)P(A)+P(B|A^{c})P(A^{c})\\=(\frac{4}{5}\times\frac{3}{5} )+(\frac{2}{5}(1- \frac{3}{5}))\\=\frac{12}{25}+\frac{4}{25}\\ =\frac{16}{25}$

• The addition rule of probability is:

        $P(Y\cup X) = P(Y) + P(X) - P(Y\cap X)$

Compute the probability that exactly one photo detector of a pair is acceptable using the addition rule of probability as follows:

$P(A\cup B) = P(A) + P(B) - P(A\cap B)\\=P(A) + P(B) - P(B|A)P(A)\\=\frac{3}{5} +\frac{16}{25} -[\frac{4}{5}\times\frac{3}{5}] \\=\frac{3}{5} +\frac{16}{25}-\frac{12}{25}\\=\frac{15+16-12}{25}\\= \frac{19}{25}$

Thus, the probability that exactly one photo detector of a pair is acceptable is $\frac{19}{25}$.