+0

# Math

-1
256
3
+484

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

#1
+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
+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)

Rom Jan 4, 2019
#2
+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
+6186
+1

well done

Rom  Jan 7, 2019