A rectangular mat has a length of 12 in. and a width of 4 in. Drawn on the mat are three circles. Each circle has a radius of 2
in. Some friends play a game throwing coins onto the mat to see who can land a coin on one of the circles.
Assuming that a coin always lands on the mat, what is the probability that a coin thrown at random will land inside one of the circles?
Use 3.14 to estimate Pi
Enter your answer, as decimal rounded to the nearest hundredth.
Please help thank you so much!
Assuming that a coin always lands on the mat, what is the probability that a coin thrown at random will land inside one of the circles?
Use 3.14 to estimate Pi
Enter your answer, as decimal rounded to the nearest hundredth.
Please help thank you so much!
2 answers:
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The question is incomplete. Here is the complete question.
The probability density function of the time to failure of an electronic component in a copier (in hours) is
for x > 0. Determine the probability that
a. A component lasts more than 3000 hours before failure.
b. A componenet fails in the interval from 1000 to 2000 hours.
c. A component fails before 1000 hours.
d. Determine the number of hours at which 10% of all components have failed.
Answer: a. P(x>3000) = 0.5
b. P(1000<x<2000) = 0.2325
c. P(x<1000) = 0.6321
d. 105.4 hours
Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.
So, probability function is given by:
P(a<x<b) =
Then, for the electronic component, probability will be:
P(a<x<b) =
P(a<x<b) =
P(a<x<b) =
a. For a component to last more than 3000 hours:
P(3000<x<∞) =
Exponential equation to the infinity tends to zero, so:
P(3000<x<∞) =
P(3000<x<∞) = 0.05
There is a probability of 5% of a component to last more than 3000 hours.
b. Probability between 1000 and 2000 hours:
P(1000<x<2000) =
P(1000<x<2000) =
P(1000<x<2000) = 0.2325
There is a probability of 23.25% of failure in that interval.
c. Probability of failing between 0 and 1000 hours:
P(0<x<1000) =
P(0<x<1000) =
P(0<x<1000) = 0.6321
There is a probability of 63.21% of failing before 1000 hours.
d. P(x) =
0.1 =
-x = -1000.ln(0.9)
x = 105.4
10% of the components will have failed at 105.4 hours.