Question 1190870
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Part (a)


Original Table<table border = "1" cellpadding = "5"><tr><td>Blood type</td><td>U.S. probability</td></tr><tr><td>A</td><td>0.42</td></tr><tr><td>B</td><td>0.11</td></tr><tr><td>AB</td><td>0.03</td></tr><tr><td>O</td><td>?</td></tr></table>
With any probability distribution, we need to have the following<ul><li>Every P(X) value is between 0 and 1, ie *[tex \large 0 \le P(X) \le 1]</li><li>All of the P(X) values add to 1 so that we account for 100% of every event possible.</li></ul>Let m be the missing probability in the table
It must add with the other probabilities to get to 1.
0.42+0.11+0.03+m = 1
0.56+m = 1
m = 1-0.56
m = 0.44
This tells us that 44% of patients in the US have type O blood.


This is what the table looks like with the missing item filled in properly<table border = "1" cellpadding = "5"><tr><td>Blood type</td><td>U.S. probability</td></tr><tr><td>A</td><td>0.42</td></tr><tr><td>B</td><td>0.11</td></tr><tr><td>AB</td><td>0.03</td></tr><tr><td>O</td><td>0.44</td></tr></table>
I'll let you check the two requirements I mentioned earlier.


Answer: 0.44


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Part (b)


Add up the probabilities for type O and type A blood
0.42+0.44 = 0.86


Answer: 0.86
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