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
| Z | P(Z) |
| 0 | 2/11 |
| 1 | 6/11 |
| 2 | 3/11 |
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)
| Z | P(Z) |
| 0 | 0.18 |
| 1 | 0.55 |
| 2 | 0.27 |
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.