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a)

X      P[X]

0       5/14

1        15/28

2       3/28

b)

The expected value is 0.75

Step-by-step explanation:

Ok, we know that out of 8 cameras, 3 are defective.

So first let's find the probability for a camera randomly selected to be defective.

This is just the quotient between the number of defective cameras and the total number of cameras.

p = 3/8

then the probability that a camera is not defective is:

q = 5/8.

Ok, now we draw 2 cameras at random from the box.

We can define X as the number of defective cameras in these two drawn, we can have 3 possible values of X.

X = 0  (neither of the cameras is defective)

X = 1 (one of the cameras is defective)

X = 2 (both of the cameras is defective).

Let's find the probabilities for each case.

X = 0.

In this case, we first draw a non-defective camera, with a probability of:

P = 5/8.

The second camera drawn must be also non-defective, but now there are 4 non-defective cameras in the box and a total of 7 cameras (because one was already drawn).

Then the probability now is:

Q = 4/7

The joint probability is the product of the two individual probabilities:

P[0] = P*Q = (5/8)*(4/7) = (5/14)

X = 1

Here we have two cases:

the first is defective and the second is non-defective

the first is non-defective and the second is defective

So we just have a factor of 2, to consider both cases

Assuming the first case

Probability of drawing first a defective camera is equal to the quotient between the number of defective cameras  and the total number of cameras:

P = 3/8

For the second draw we want to get a non-defective camera, here the probability is equal to the number of non-defective cameras remaining (5) and the total number of cameras (7, because we drawn one)

Q = 5/7

The joint probability, taking in account the permutation, is

P[1] = 2*P*Q = 2*(3/8)*(5/7) = (15/28)

finally, for X = 2

This is the case where we draw two defective cameras, we can use a similar approach as the one used in the first case:

For the first camera:

P = 3/8

For the second camera:

Q = 2/7

Joint probability:

P[2] = (3/8)*(2/7) = 3/28

then we have the table:

X      P[X]

0       5/14

1        15/28

2       3/28

b)

The expected value for an event that has the outcomes:

{x₁, x₂, ..., xₙ}

Each one with the correspondent probability

{p₁, p₂, ..., pₙ}

is defined as:

EV = x₁*p₁ + x₂*p₂ + ... + xₙ*pₙ

Then in our case, the expected value is just:

EV = 0*P[0] + 1*P[1] + 2*P[2]

EV = 0  + 15/28  + 2*3/28

EV = (15 + 6)/28 = 21/28 = 0.75

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: