Question

Let A and B be two events associated with an experiment such that 4.0)A(P= and 7.0)BA(P=∪. Compute )B(P when i) A and B are mutually exclusive. ii) A and B are independent.

Answer 1

Answer on Question #65070 – Math – Statistics and Probability

Question

Let $A$ and $B$ be two events associated with an experiment such that

$$P(A) = 0.4$$

and

$$P(A \cup B) = 0.7.$$

Compute $P(B)$ when

i) $A$ and $B$ are mutually exclusive;

ii) $A$ and $B$ are independent.

Solution

i) Recall that, for any two events $A$ and $B$, we have the inclusion-exclusion formula [1, p.29] for probabilities:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$

Assume that events $A$ and $B$ are mutually exclusive [1, p.24].

Then by definition,

$$P(A \cap B) = 0.$$

Hence the following formula can be obtained:

$$P(B) = P(A \cup B) - P(A).$$

So,

$$P(B) = 0.7 - 0.4 = 0.3.$$

ii) Again, we can apply the inclusion-exclusion formula [1, p.29]:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$

By the definition of independence, two events $A$ and $B$ are independent [1, p.79] if and only if

$$P(A \cap B) = P(A)P(B).$$

Thus,

$$P(A \cup B) = P(A) + P(B) - P(A)P(B).$$

Now, express this formula in terms of $P(B)$, bearing in mind that $P(A) \neq 1$:

$$P(B) = \frac{P(A \cup B) - P(A)}{1 - P(A)}.$$

Finally, we can compute the answer:

$$P(B) = (0.7 - 0.4)/(1 - 0.4) = 0.3/0.6 = 0.5.$$

References:

1. Sheldon M. Ross. *A First Course in Probability*. 8th edition.

2. Richard N. Aufmann, Joanne S. Lockwood, Richard D. Nation, Daniel K. Clegg. *Mathematical Excursions*. 2nd edition.

3. Charles Henry Brase, Corrinne Pellillo Brase. *Understandable Statistics: Concepts and Methods*. 10th edition.

Answer:

i) $P(B) = 0.3$

ii) $P(B) = 0.5$

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