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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

i dont know sorry

step-by-step explanation:

question: how many people can she serve?

answer: she can serve 16.66666667 people. so she can only give 16 people a complete serving.

question: does she have enough ice cream for her guests?

answer: no. explanation: she only has enough for 16 people to have a full serving. however, she can give a 17th person a partial serving.