9. In order for a circuit board to work, seven identical chips must be in workin

Question

9. In order for a circuit board to work, seven identical chips must be in working order. To improve reliability, an additional chip is included in the board, and the design allows it to (a) the probability p, that the board is working in tems of the probability p thai an (b) Suppose that n circuit boards are operated in parallel, and that we require a 99.9% replace any of the seven other chips when they fail individual chip is working. probability that at least one board is working. How many boards are needed?

Explanation / Answer

The number of working chips follows a binomial diustribution ~ Binom(8,p)

(a) For the circuit board to work, at least seven chips must be working at any given time,

So P(board is working) = P(7 chips are working) + P(8 chips are working)

pb =P(board is working) = P(7 chips are working) + P(8 chips are working) = C[8,7] x p7 x (1-p) + C[8,8] x p8 x (1-p)0

= 8*p7*(1-p) + p8

(b) We require the probability that at least one of the boards is working to be at least 99.9%, i.e., a probability > 0.999

So the probability that none of the boards is working should be less than 1-0.999 = 0.001

Probability that none of the boards is working = (1-pb)n

So we have, (1-pb)n 0.001

log (1-pb)n log (0.001)

n log (1-pb) log (0.001)

n log (0.001) / log (1-pb)

The inequality changes from to since log (1-pb) is a negative number, 1-pb being less than 1.

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