# Math Help - Random Variables and Prob Distributions

1. ## Random Variables and Prob Distributions

1. A shipment of 7 television sets contains 2 defectve sets. A hotel makes a random purchase of 3 sets. If x is the number of defective sets purchased by the hotel, find the probability distribution of X.

my solution:
x can take {0,1,2} since only 2 defective sets.
sample space = 7choose 3 = 70

f(x) = P(X=0) = (2 choose 0 ) * (5choose2) / (7 choose 3) = 20/70= 2/7
P(X=1) ?
P(X=2) ?

i am stuck.

2. The time to failure in hours of an important pc of electronic equpment used in a manufactured DVD player has the density function

f(x) = 1/2000 exp (-x/2000) , x>=0
0, x<0

2a) Find F (x)

2b) Determine the prob that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced.

2c) Determine the probability that the component fails before 2000 hours.

many thanks !

2. Q1.
Firstly 7C3 = 35 not 70.
p(x=0)= 2C0 x 5C3 / 7C3 (ie 0 defectives out of 2, and 3 good ones out of 5)
p(x=1) = 2C1 x 5C3 / 7C3 (ie 1 defective out of 2 and 2 good ones out of 5)
p(x=2) = ....you can do that one!

(Good idea to check when you finish that they all add up to 1.)

3. Have you had a go at Q2. It involves integrating and finding area under the curve. A graph will help you think it through.

4. Originally Posted by hazel
[snip]
f(x) = P(X=0) = (2 choose 0 ) * (5choose2) / (7 choose 3) = 20/70= 2/7
P(X=1) ?
P(X=2) ?

i am stuck.

2. The time to failure in hours of an important pc of electronic equpment used in a manufactured DVD player has the density function

f(x) = 1/2000 exp (-x/2000) , x>=0
0, x<0

2a) Find F (x)

2b) Determine the prob that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced.

2c) Determine the probability that the component fails before 2000 hours.

many thanks !
Have you been taught how to use a probability density function (pdf) to calculate probabilities ....? Please post all the work you've done and state where you get stuck.

5. ## Ok for qn 1

Originally Posted by Debsta
Q1.
Firstly 7C3 = 35 not 70.
p(x=0)= 2C0 x 5C3 / 7C3 (ie 0 defectives out of 2, and 3 good ones out of 5)
p(x=1) = 2C1 x 5C3 / 7C3 (ie 1 defective out of 2 and 2 good ones out of 5)
p(x=2) = ....you can do that one!

(Good idea to check when you finish that they all add up to 1.)
Oh yes, i got it now. Thanks !

6. So does it mean i do an integration of the expression 1/2000 exp (-x/2000) ?
but the thing is i don't know the range.
what i don't understand is - x>=0 and then for 0 x < 0.

7. Originally Posted by hazel
So does it mean i do an integration of the expression 1/2000 exp (-x/2000) ?
but the thing is i don't know the range.
what i don't understand is - x>=0 and then for 0 x < 0.
You are told that f(x) = 0 for x < 0 and that it's equal to 1/2000 exp (-x/2000) for $x \geq 0$.

8. So meaning just integrate this function and then substitute x= 0 to get the answer ?

9. Originally Posted by hazel
So meaning just integrate this function and then substitute x= 0 to get the answer ?
$\Pr(X > a) = \int_a^{+\infty} f(x) \, dx$.

$\Pr(X < a) = \int^a_{0} f(x) \, dx$ where f(x) is the rule that applies for when x > 0.

These are things you're meant to know.

10. Originally Posted by mr fantastic
$\Pr(X > a) = \int_a^{+\infty} f(x) \, dx$.

$\Pr(X < a) = \int^a_{0} f(x) \, dx$ where f(x) is the rule that applies for when x > 0.

These are things you're meant to know.
ok thanks.