We have a system of 10 satellites. The probability that a satellite works on a maintenance day is 0.85 and the probability that it works on a non-maintenance day is 0.95. Maintenance days are 30% of the days.
The systems works if at least 8 of the 10 satellites work. Find the probability that the systems works on a random day.
- Solution 1
P(W|M) = 0.85, P(W|M') = 0.95, P(W) = 0.3, P(W') = 1 – P(W) = 0.7
The number of working satellites on a day is described by a binomial distribution. For the system to work we need the probability that work exactly 8 or exactly 9 or exactly 10 satellites.
The probability that the system works on a maintenance day is described by a binomial distribution:
$$P(S|M) = 10C8 * P(W|M)^8 * (1-P(W|M))^2 + 10C9 * P(W|M)^9 * (1-P(W|M)) + 10C10 * P(W|M)^{10} = 45*0.85^8*0.15^2 + 10*0.85^9*0.15 + 0.85^{10} = 0.8202 $$
The probability that the system works on a non-maintenance day is described by a binomial distribution:
$$P(S|M') = 10C8 * P(W|M')^8 * (1-P(W|M'))^2 + 10C9 * P(W|M')^9 * (1-P(W|M')) + 10C10 * P(W|M')^{10} = 45*0.95^8*0.05^2 + 10*0.95^9*0.05 + 0.95^{10} = 0.9885 $$
The probability that the system works on a random day is:
P(S) = P(S|M)*P(M) + P(S|M')*P(M') = 0.8202 * 0.3 + 0.9885 * 0.7 = 0.9380
- Solution 2
The probability that a single satellite works on a random day is:
P(W) = P(W|M)*P(M) + P(W|M')*P(M') = 0.85 * 0.3 + 0.95 * 0.7 = 0.92
The probability that the system works on a random day is:
$$P(S) = 10C8 * P(W)^8 * (1-P(W))^2 + 10C9 * P(W)^9 * (1-P(W)) + 10C10 * P(W)^{10} = 45*0.92^8*0.08^2 + 10*0.92^9*0.08 + 0.92^{10} = 0.9599 $$
According to my teacher's answer the first solution is correct. Where is my mistake with the second solution?
Best Answer
The events (one for each of the satellites) are not independent. Knowing that the first works is evidence that it is a non-maintenance day, which therefore increases the probability that the second satellite works.
To illustrate the problem, consider an extreme example: "you are are tossing an unfair coin twice. On "good days" you only get Heads, on "bad days" you only get Tails. Exactly half of days are good. What is the probability that you get one Head and one Tail on a random day?"
Of course, the answer is $0$ since the first toss yielding $H$ (resp. $T$) means the second is certain to come up $H$ (resp. $T$). But your method gives a non-zero answer. Indeed, the probability that a given toss on a random day is $H$ is $\frac 12$ (clearly) so each individual toss is as it would be for a fair coin, and your method then gives $\frac 12$ in the end.