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Is this part right for (b)

P[losing] = 0.7

P[playing away] = 0.15

P[losing and playing away] = 0.1

P[losing] P[playing away] =(0.7)(0.15) = 0.105

P[losing and playing away] ≠ P[losing] P[playing away]

Dependent

 Jan 4, 2019
edited by awsometrunt14  Jan 7, 2019

Best Answer 

 #1
avatar+6186 
+3

\(P[\text{winning}]=0.25\\ P[\text{playing at home}]=0.8\\ P[\text{winning AND playing at home}] = 0.2\\ P[\text{winning}]P[\text{playing at home}] = (0.25)(0.8) = 0.2\\P[\text{winning AND playing at home}]= P[\text{winning}]P[\text{playing at home}] \\ \text{so the two events are independent}\)

 

now that you've been shown the method apply it to (b)

 Jan 4, 2019
 #1
avatar+6186 
+3
Best Answer

\(P[\text{winning}]=0.25\\ P[\text{playing at home}]=0.8\\ P[\text{winning AND playing at home}] = 0.2\\ P[\text{winning}]P[\text{playing at home}] = (0.25)(0.8) = 0.2\\P[\text{winning AND playing at home}]= P[\text{winning}]P[\text{playing at home}] \\ \text{so the two events are independent}\)

 

now that you've been shown the method apply it to (b)

Rom Jan 4, 2019
 #2
avatar+484 
+1

Is this right?

P[losing] = 0.7

P[playing away] = 0.15

P[losing and playing away] = 0.1

P[losing] P[playing away] =(0.7)(0.15) = 0.105

P[losing and playing away] ≠ P[losing] P[playing away]

Dependent

awsometrunt14  Jan 7, 2019
 #3
avatar+6186 
+1

well done cool

Rom  Jan 7, 2019

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