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Iteru [2.4K]
3 years ago
6

Players on a softball team are randomly given a uniform with a two-digit number on the back. The numbers range from 10 to 99. Wh

at is the probability that the number on the uniform will end in 0?
Mathematics
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

1/10 or 10%

Step-by-step explanation:

99 possible numbers starting from 10

1,2,3,4,5,6,7,8,9 does not count

99-9=90

90 numbers total

possible numbers with a 0: 9

9/90 = 1/10 = 10%

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30% less of 180<br><br> ......................................................
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126

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The source of the error when a computer system fails are either the disk drive, the computer memory, or the operating system. Ve
faltersainse [42]

Answer:

The probability of an operating system error given that a failure has ocurred is P(O/E)= 0.19

Step-by-step explanation:

Hello!

There are three different sources of error when a computer system fails:

D: disk drive error. ⇒ P(D)= 0.50

M: computer memory error. ⇒P(M)= 0.20

O: operating systems error. ⇒ P(O)= 0.30

According to the component perdormance standard:

The probability of failure "E", given that there is a disk drive error is P(E/D)= 0.40

The probability of failure "E", given that there is a computer memory error is P(E/M)= 0.6

The probability of failure "E", given that there is a operating system error is P(E/O)= 0.25

You need to calculate the probability of an operating system error given that a failure has ocurred, symbolically:

P(O/E)

P(O/E)= \frac{P(OnE)}{P(E)}

To reach the probability of the marginal E you have to add all intersections between the event "E" and the events "D", "M" and "O"

       D           M             O       Total

E:  P(E∩D); P(E∩M); P(E∩O);  P(E)

Using the formula of the conditional probability I'll clear all three intersections using the known probabilities:

General formula: P(A/B)= \frac{P(AnB)}{P(B)} ⇒ P(AnB)= P(A/B)*P(B)

P(E∩D)= P(E/D)*P(D)= 0.4*0.5= 0.2

P(E∩M)= P(E/M)*P(M)= 0.6*0.2= 0.12

P(E∩O)= P(E/O)*P(O)= 0.25*0.3= 0.075

P(E)= P(E∩D) + P(E∩M) + P(E∩O)= 0.2+0.12+0.075= 0.395

P(O/E)= \frac{P(OnE)}{P(E)} = \frac{0.075}{0.395}= 0.189 = 0.19

I hope it helps!

7 0
3 years ago
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