Question 1189844: Construct the probability distribution for the random variables described in each of the following situations. Draw the corresponding histogram for each probability distribution.
1. Two balls are drawn in succession without replacement from an urn containing 5 red balls and 6 blue balls. Let Z be the random vatiable representing the number of blue balls. Find the values of the random variable Z.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
We're selecting two balls without replacement.
There are 5 red and 6 blue, giving 11 total.
We have three possible outcomes:
A) There are 0 blue selected
B) There is exactly 1 blue selected
C) There are exactly 2 blue selected
Each scenario corresponds to one histogram bar or rectangle. See below.
Let's calculate the probability for scenario A.
5 red out of 11 total gives 5/11 the probability of getting red
After picking that red, we have 4 left out of 10 total (subtract 1 from each previous value). That gives 4/10 as the next probability of getting red.
The probability of two red in a row is
(5/11)*(4/10) = 20/110 = 2/11
2/11 = 0.18 approximately
There's roughly an 18% chance that scenario A happens
Because P(2 red) = 2/11, this means P(0 blue) = 2/11 also.
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Now onto scenario B
P(1st red, 2nd blue) = (5/11)*(6/10) = 30/110 = 3/11
P(1st blue, 2nd red) = (6/11)*(5/10) = 30/110 = 3/11
P(1 red, 1 blue in either order) = 2*3/11 = 6/11
P(exactly 1 blue) = 6/11
6/11 = 0.55 approximately
There's roughly a 55% chance that scenario B happens
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Lastly, scenario C
P(2 blue) = P(blue)*P(blue) = (6/11)*(5/10) = 30/110 = 3/11
3/11 = 0.27 approximately
There's roughly a 27% chance that scenario C happens
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We calculated that
P(0 blue) = 2/11 = 0.18
P(1 blue) = 6/11 = 0.55
P(2 blue) = 3/11 = 0.27
Notice that the three probability values add to 1, which represents 100% of all possible cases.
0.18+0.55+0.27 = 1 = 100%
This is what the probability distribution could look like in table format
Z = number of blue selected
It's similar to a x,y table for any type of function. In this case our input is Z = the number of blue selected, and the output of the function is P(Z) = the probability of getting that Z value.
Here's an equivalent table but each fraction is replaced with its approximate decimal counterpart (eg: 2/11 = 0.18)
Below is the histogram. Note the bars do not have a gap between them.
I used the decimal values to get each bar height, but you could use the fraction form if you wanted.
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