Answer
a) $0.57$
b)$0.95$
c)$0.83$
d)$0.38$
e)$0.29$
f)$0.05$
g)$0.78$
h)$0.71$
Work Step by Step
These events are mutually exclusive hence I can obtain the required results by simply adding the individual probabilities.
We know that probability$=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}.$
a) P(1 or 2 TV sets)=$P(1)+P(2)=0.24+0.33=0.57$
b) P(1 or more TV sets)=$P(1)+P(2)+P(3)+P(\geq4)=0.24+0.33+0.21+0.17=0.95$
c)P(3 or fewer TV sets)=$P(0)+P(1)+P(2)+P(3)=0.05+0.24+0.33+0.21=0.83$
d)P(3 or more TV sets)=$P(3)+P(\geq4)=0.21+0.17=0.38$
e)P(fewer than 2 TV sets)=$P(0)+P(1)=0.05+0.24=0.29$
f) P(fewer than 1 TV sets)=$P(0)=0.05$
g)P(1, 2 or 3 TV sets)=$P(1)+P(2)+P(3)=0.24+0.33+0.21=0.78$
h)P(2 or more TV sets)=$P(2)+P(3)+P(\geq4)=0.33+0.21+0.17=0.71$
