In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour T.V., a black and white T.V. or both kinds of T.V. The probability that a family own either any one or both kinds of T.V. sets is

Options

A. 0.87
B. 0.93
C. 0.85
D. 0.76

Answer

To solve the problem, we need to find the probability that a family owns either a color TV, a black and white TV, or both. We can denote the events as follows:

- Let \( E_1 \) be the event that a family owns a color TV.
- Let \( E_2 \) be the event that a family owns a black and white TV.

From the question, we have the following probabilities:

- \( P(E_1) = 0.87 \) (the probability that a family owns a color TV)
- \( P(E_2) = 0.36 \) (the probability that a family owns a black and white TV)
- \( P(E_1 \cap E_2) = 0.30 \) (the probability that a family owns both types of TVs)

We need to find \( P(E_1 \cup E_2) \), which is the probability that a family owns either a color TV, a black and white TV, or both. We can use the formula for the union of two events:

\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \]

Now, we can substitute the values we have:

\[ P(E_1 \cup E_2) = 0.87 + 0.36 - 0.30 \]

Calculating this step-by-step:

1. Add \( P(E_1) \) and \( P(E_2) \):
\[ 0.87 + 0.36 = 1.23 \]

2. Subtract \( P(E_1 \cap E_2) \):
\[ 1.23 - 0.30 = 0.93 \]

Thus, the probability that a family owns either a color TV, a black and white TV, or both is:

\[ P(E_1 \cup E_2) = 0.93 \]

### Final Answer:
The probability that a family owns either any one or both kinds of TV sets is **0.93**.