Question

4.Suppose that 80 percent of all families own television set. If 10 families are interviewed at
random, find the probability that at most three families own a television set.
​

Answer 1

Answer:

3/10

Step-by-step explanation:

probalility of having at most 3 families a television set of own =3/10

Answer 2

The required probability is 0.00086.

Step-by-step explanation:

Consider the provided information.

80 percent of all families own television set.

Thus, p = 0.8 and q =  1-0.8 = 0.2

10 families are interviewed at  random,

Thus, n = 10

Now we need to find the probability that at most three families own a television set.

$P(X\leq3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X\leq3)=^{10}C_0(0.8)^0(0.2)^{10}+^{10}C_1(0.8)^1(0.2)^9+^{10}C_2(0.8)^2(0.2)^8+^{10}C_3(0.8)^3(0.2)^7$

$P(X\leq3)=0.0008643584$

Hence, the required probability is 0.00086.

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