Mathematics, 12.03.2020 06:08 brialevy2283

A large bank vault has several automatic burglar alarms. The probability is 0.55 that a single alarm will detect a burglar. (a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?

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Answer from: laycihyatt11

6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

Step-by-step explanation:

For each alarm, there are only two possible outcomes. Either a theft is detected, or is not. The probability of an alarm detecting a theft is independent of others alarms. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.55 that a single alarm will detect a burglar.

This means that p = 0.55

(a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?

The probability that no alarms detect a burglar is P(X = 0). Either no alarms detect, or at least one does. We need to have a 99% probability that at least one does. So we need P(X = 0) = 1 - 0.99 = 0.01.

We do by trial and error, find n until P(X = 0) <= 0.01.

n = 1

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1,0}.(0.55)^{0}.(0.45)^{1} = 0.45$

n = 2

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.55)^{0}.(0.45)^{2} = 0.2025$

n = 3

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$

n = 4

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.55)^{0}.(0.45)^{4} = 0.0410$

n = 5

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.55)^{0}.(0.45)^{5} = 0.0185$

n = 6

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.55)^{0}.(0.45)^{6} = 0.0083$

So we need 6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

Answer from: Quest

what is the original number? you forgot to add it.

Answer from: Quest

Answer

1 number

explanation

if we use a graphing calculator, there is only one such number $c\in[0.5,0.75]$ that satisfies the conclusion of the mean value theorem, that is, a number $c\in[0.5,0.75]$ such that

$f'(c) = \dfrac{f(0.75) - f(0.5)}{0.75-0.5},$

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

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