Math

QuestionGiven a 6%6\% disk drive failure rate, find the probability of having at least one working drive with 2 and 3 drives.

Studdy Solution

STEP 1

Assumptions1. The rate of disk drive failure in a year is6%. . The data is stored on independent hard disk drives.
3. The probability of a disk drive not failing is1 minus the probability of it failing.
4. The probability of at least one disk drive not failing is1 minus the probability of all disk drives failing.

STEP 2

First, we need to find the probability of a disk drive not failing in a year. We can do this by subtracting the rate of disk drive failure from1.
(not failing)=1(failing)(\text{not failing}) =1 -(\text{failing})

STEP 3

Now, plug in the given value for the rate of disk drive failure to calculate the probability of a disk drive not failing.
(not failing)=16%(\text{not failing}) =1 -6\%

STEP 4

Convert the percentage to a decimal value.
6%=0.066\% =0.06(not failing)=10.06(\text{not failing}) =1 -0.06

STEP 5

Calculate the probability of a disk drive not failing.
(not failing)=10.06=0.94(\text{not failing}) =1 -0.06 =0.94

STEP 6

Now that we have the probability of a disk drive not failing, we can find the probability of at least one disk drive not failing when there are two disk drives. This is1 minus the probability of both disk drives failing.
(at least one not failing)=1(both failing)(\text{at least one not failing}) =1 -(\text{both failing})

STEP 7

The probability of both disk drives failing is the product of their individual probabilities of failing.
(both failing)=(failing)×(failing)(\text{both failing}) =(\text{failing}) \times(\text{failing})

STEP 8

Plug in the values for the probability of failing to calculate the probability of both disk drives failing.
(both failing)=0.06×0.06(\text{both failing}) =0.06 \times0.06

STEP 9

Calculate the probability of both disk drives failing.
(both failing)=.06×.06=.0036(\text{both failing}) =.06 \times.06 =.0036

STEP 10

Now that we have the probability of both disk drives failing, we can subtract this from to find the probability of at least one disk drive not failing.
(at least one not failing)=0.0036(\text{at least one not failing}) = -0.0036

STEP 11

Calculate the probability of at least one disk drive not failing.
(at least one not failing)=0.0036=0.9964(\text{at least one not failing}) = -0.0036 =0.9964With two hard disk drives, the probability that catastrophe can be avoided is0.9964.

STEP 12

Now, let's find the probability of at least one disk drive not failing when there are three disk drives. This is minus the probability of all three disk drives failing.
(at least one not failing)=(all three failing)(\text{at least one not failing}) = -(\text{all three failing})

STEP 13

The probability of all three disk drives failing is the product of their individual probabilities of failing.
(all three failing)=(failing)×(failing)×(failing)(\text{all three failing}) =(\text{failing}) \times(\text{failing}) \times(\text{failing})

STEP 14

Plug in the values for the probability of failing to calculate the probability of all three disk drives failing.
(all three failing)=0.06×0.06×0.06(\text{all three failing}) =0.06 \times0.06 \times0.06

STEP 15

Calculate the probability of all three disk drives failing.
(all three failing)=0.06×0.06×0.06=0.000216(\text{all three failing}) =0.06 \times0.06 \times0.06 =0.000216

STEP 16

Now that we have the probability of all three disk drives failing, we can subtract this from to find the probability of at least one disk drive not failing.
(at least one not failing)=0.000216(\text{at least one not failing}) = -0.000216

STEP 17

Calculate the probability of at least one disk drive not failing.
(at least one not failing)=0.000216=0.999784(\text{at least one not failing}) = -0.000216 =0.999784With three hard disk drives, the probability that catastrophe can be avoided is0.999784.

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