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
