Question

a burgular alarm system has six fail-safe components. the probability of each failing is 0.05. Find these probabilities a)Exactly three will fail b) Fewer than two will fail. c) None will fail. d) Compare the answers from a,b,and c and explain why the results are reasonable

Answer 1

set up a binomial probability

Answer 2

this is but a few ways the parts can fail i think

\begin{array}c
a&b&c&d&e&f\\
0&0&0&x&x&x\\
0&0&x&0&x&x\\
0&x&0&0&x&x\\
x&0&0&0&x&x\\
0&0&x&x&x&0\\
0&x&0&x&x&0\\
x&0&0&x&x&0\\
0&0&x&x&x&0\\
0&x&x&x&0&0\\
x&0&x&x&0&0\\
x&x&x&0&0&0\\
\end{array}

Answer 3

Let's enumerate the fail-safe components, with numbers from 1 to 6. If we choose three of those components, let's say 3, 4 and 6, the probability of the three failing is \[0.05^3 (1-0.05)^3.\]
We can choose the three components that will fail in more than one way. The probability of 4, 5 and 6 failing is again \[0.05^3(1-0.05)^3\]and because these are disjoint events, the probability of 3, 4 and 6 failing or 4, 5 and 6 failing is the sum of the two previous numbers. If we want the probability of *any* three components failing, we must multiply the first number by the amount of ways we can choose three numbers out of six. This is a binomial coefficient:
\[\binom{6}{3}\]

So the answer to a) is:

\[\binom{6}{3} 0.05^3 (1-0.05)^3 = \frac{6859}{3200000}.\]

The next parts are very similar.

Answer 4

lol; yeah, like that

Answer 5

ok thank you