Question 1189844
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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<table border = "1" cellpadding = "5"><tr><td>Z</td><td>P(Z)</td></tr><tr><td>0</td><td>2/11</td></tr><tr><td>1</td><td>6/11</td></tr><tr><td>2</td><td>3/11</td></tr></table>
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)<table border = "1" cellpadding = "5"><tr><td>Z</td><td>P(Z)</td></tr><tr><td>0</td><td>0.18</td></tr><tr><td>1</td><td>0.55</td></tr><tr><td>2</td><td>0.27</td></tr></table>



Below is the histogram. Note the bars do not have a gap between them.
<img width = "50%" src = "https://i.imgur.com/Jw0KZYF.png">
I used the decimal values to get each bar height, but you could use the fraction form if you wanted. 
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