Question
2) here we need to find P(X4)=P(X=4)+P(X=3)+ P(X=2)+P(X=1)+P(X=0) =(415)(0.20)4(0.80)11++(015)(0.20)0(0.80)15 =0.188+0.250+0.231+0.132 +0.035= 0.836 (round 3 decimal) This came from a solution - what does the ++
2) here we need to find
P(X4)=P(X=4)+P(X=3)+
P(X=2)+P(X=1)+P(X=0)
=(415)(0.20)4(0.80)11++(015)(0.20)0(0.80)15
=0.188+0.250+0.231+0.132
+0.035= 0.836 (round 3 decimal)
This came from a solution - what does the ++ in the middle of the solution mean?
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Elementary Linear Algebra
Authors: Ron Larson, David C Falvo
6th Edition
130517240X, 9781305172401
