Question 1123857: A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.
For (a) and (b), use x/y. For example: 9/10
Compute P(T).
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Compute P(T|F).
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Are T and F independent? Yes or No.
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Are F and S mutually exclusive? Yes or No.
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Are F and S independent? Yes or No.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Define the following:
T = event of getting the white ball twice
F = event of getting the white ball first
S = event of getting the white ball second
For any selection, the chances of picking a whitball is 1/2 since there is 1 white ball out of two total. Replacement is happening meaning that the second selection probability of getting white is not affected by the first selection. Each selection is independent of one another. Therefore, we can multiply the probabilities like so
P(T) = P(getting white the first time)*P(getting white the second time)
P(T) = (1/2)*(1/2)
P(T) = 1/4 is the answer to the first part
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Let's label the balls as W and R for white and red respectively. We have this sample space:
WW
WR
RW
RR
to represent all of the possible outcomes. As you can see, WW only happens once out of 4 times, which helps back up the previous answer 1/4.
Also, we see that W happens first exactly two out of the four cases (WW and WR), so we have 2/4 = 1/2 as the probability of picking white first. This means P(F) = 1/2
The probability of picking white twice and picking white first is the same as picking white twice. Why? Because if you picked white twice then white has to happen first, or else it wouldn't make any sense. So P(T and F) = P(T). We say that event T is a subset of event F.
So,
P(T|F) = P(T and F)/P(F)
P(T|F) = P(T)/P(F)
P(T|F) = (1/4) divided by (1/2)
P(T|F) = (1/4)*(2/1)
P(T|F) = 2/4
P(T|F) = 1/2
What does this mean in the sample space context? Well we have this sample space
WW
WR
RW
RR
and we know for a fact that we got W first based on the fact that event F is given. So we can reduce the sample space to just this
WW
WR
and we see that WW happens once out of two possible cases. This is why the answer to this part is 1/2
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To say F and S are mutually exclusive is the same as saying "F and S cannot possibly happen at the same time". Let's see if that is the case or not.
Recall that we let,
F = event of getting a white ball first
S = event of getting a white ball second
But the scenario WW is possible so this is proof that F and S are not mutually exclusive.
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Are F and S independent? To answer that question, we need to see if event F alters event S in any way (and vice versa)
If we get a white ball first, then P(S) is going to stay the same because we put the ball back into the box. Each event is independent of one another. So F is the first event and S is the second event. That might be a good way to think of it.
In short, yes F and S are independent events
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A more formal proof would be to show that P(F and S) = P(F)*P(S) is true
P(F and S) = 1/4 since we have one WW event out of four
P(F) = 1/2 found earlier
P(S) = 1/2 which is similar to P(F)
P(F)*P(S) = (1/2)*(1/2) = 1/4
Both sides have the result of 1/4, which proves P(F and S) = P(F)*P(S) to be true; hence the events F and S are indeed independent
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