In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Home » In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Solution:

Event that a family owns colour television $=E_{1}$
Event that the family owns black and white television $=E_{2}$
Provided that $P\left(E_{1}\right)=0.87$
$P\left(E_{2}\right)=0.36$ and $P\left(E_{1} \cap E_{2}\right)=0.30$
We now need to find the probability that a family owns either anyone or both kinds of sets.
Using the General Addition Rule, we get
$\begin{array}{l} P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ \therefore P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right) \\ =0.87+0.36-0.30 \\ =1.23-0.30 \\ =0.93 \end{array}$
As a result, the required probability is $0.93$