Mathematics, 28.07.2019 07:00, ansley81

Among 72647264 cases of heart pacemaker malfunctions, 386386 were found to be caused by firmware, which is software programmed into the device. if the firmware is tested in 33 different pacemakers randomly selected from this batch of 72647264 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted

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Among 7264 cases of heart pacemaker malfunctions, 386 were found to be caused by firmware, which is software programmed into the device. if the firmware is tested in 3 different pacemakers randomly selected from this batch of 7264 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted
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This problem properly belong to the hypergeometric distribution (select with no replacements) but because of the small size of the sample, binomial distribution (constant probability) would model the problem quite closely.

We will attempt both methods.

1. Hypergeometric distribution:
Use hypergeometric distribution where:a=number of defective items selectedA=total number of defective itemsb=number of non-defective items selectedB=total number of non-defective itemsThen $P(a,b)=\frac{C(A,a)C(B,b)}{C(A+B,a+b)}$ where $C(n,r)=\frac{n!}{(n!(n-r)!)}$ = combination of r items selected from n,A+B=total number of items = 7264a+b=number of items selected=3
A=386
a=0
b=3
B=7264-386=6878

$P(0,3)=\frac{C(386,0)C(6878,3)}{C(7264,3)}$
$=\frac{1*54205806876}{63855288864}$
=0.848885 (to the sixth decimal place)

2. Binomial distribution
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$

In the present case, it is used as an approximation because the selection is without replacement, so probability, p, varies with the three trials.

Here
p=386/7264 (approximately)
n=7264
x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(7264,0)(386/7264)^0(1-(386/7264))^{3-0}$
=1*1*(40672093519/47911251968)

=0.848905

The error in the binomial approximation is minimal, namely
Error=1-0.848905/0.848885
=+0.00232%

So the probability that the entire batch will be accepted is 0.8489

answered: Guest

Suredude why not right

answered: Guest

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

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Among 72647264 cases of heart pacemaker malfunctions, 386386 were found to be caused by firmware, wh...