I am a little bit confused by an exercise.
Given a system consisting of 6 elements with reliabilities: $p_2=p_3=p_4=p_5=1/2, p_1 = 2/3, p_6 = 3/4.$ Find a probability of exactly two elements failure in parallel connection.
My try was to simply apply Binomial probability formula: $C_4^2 (\frac{1}{2})^2 (1 – \frac{1}{2})^{4 – 2}$ and multiply it by $p_1$ which gives me a probability of 1/4. But the authors consider 2/5 as a correct answer.
Thank You!
Best Answer
UPD: Oops :( I missed a huge detail. "Given that the system works". So I get the correct answer of 2/5.
Let $P(C)$ - probability that system works, $P(A)$ - probability that 2 out of 4 parallel elements failed.
Then $P(C) = p_1 * (1 - \frac{1}{16})*p_6$. (1 - 1/16 - is a probability of at least 1 parallel element works). $P(C) = \frac{15}{32}$.
$P(A) = {4 \choose 2} \frac{1}{2}^2 \frac{1}{2}^2 = \frac{3}{8}$.
Now we need to find a probability of P(A|C).
$P(A|C) = \frac{P(AC)}{P(C)} = \frac{P(A)P(C|A)}{P(C)} = \frac{\frac{3}{8}\frac{6}{12}}{15/32} = \frac{2}{5}.$