Answer:
a.[tex]a. \ \frac{7}{25}[/tex]
[tex]b.\ \ \ P(D_1D_2)=\frac{6}{25}[/tex]
Explanation:
a. Find the probability that exactly one photo detector of a pair is acceptable:
Let [tex]A_i=i^{th}[/tex] photo is accepted and the probability [tex]D_i=i^{th}[/tex] is defected.
Therefore:
[tex]P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}[/tex]
#The probability of exactly one photo detector of a pair is accepted is 7/25
b.Find the probability that both photo detectors in a pair are defective,P(D1D2):
[tex]P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}[/tex]
Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25
