Question

If P(A)= 0.3 , P (B)= 0.2, and P(A and B)= 0.1 , determine the following probabilities: (a) P( A' ) (b) P (A U B) (c) P(A' INTERSECT B) (d) P [ (A U B)' ] (e) P (A' U B)'

Answer 1

Concepts and reason

Event: The collection or the set of outcomes in an experiment is called as an event.

Complementary events: If an event A is defined, then complement of event A is not occurring of event A.

Union of two events: The set of the outcomes that belong to either the two events or any of the two events is called as union of two events.

Intersection of two events: The set of the outcomes that belong to both the two events s is called as intersection of two events.

Fundamentals

The probability value for complementary event (A) is, $P\left( {A'} \right) = 1 - P\left( A \right)$

The addition rule for the probability is, $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

Some of the other formulas for probability are,

$\begin{array}{c}\\P\left( {A' \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right)\\\\P\left( {A \cup B} \right)' = 1 - P\left( {A \cup B} \right)\\\end{array}$

$\begin{array}{c}\\P\left( {A' \cup B} \right)' = 1 - P\left( {A' \cup B} \right)\\\\ = 1 - \left[ {P\left( {A'} \right) + P\left( B \right) - P\left( {A' \cap B} \right)} \right]\\\end{array}$

(a)

The value for $P\left( {A'} \right)$ is obtained as shown below:

From the given information, $P\left( A \right) = 0.3$

The probability of complement for A is,

$\begin{array}{c}\\P\left( {A'} \right) = 1 - P\left( A \right)\\\\ = 1 - 0.3\\\\ = 0.7\\\end{array}$

(b)

The value for $P\left( {A \cup B} \right)$ is obtained as shown below:

From the given information, $P\left( A \right) = 0.3\,\,,\,\,P\left( B \right) = 0.2\,\,,P\left( {A\, \cap \,B} \right) = 0.1$

The required probability is,

$\begin{array}{c}\\P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\\\ = 0.3 + 0.2 - 0.1\\\\ = 0.5 - 0.1\\\\ = 0.4\\\end{array}$

(c)

The value for $P\left( {A' \cap B} \right)$ is obtained as shown below:

From information given, $P\left( B \right) = 0.2\,\,,P\left( {A\, \cap \,B} \right) = 0.1$

The required probability is,

$\begin{array}{c}\\P\left( {A' \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right)\\\\ = 0.2 - 0.1\\\\ = 0.1\\\end{array}$

(d)

The value for $P\left( {A \cup B} \right)'$ is obtained as shown below:

From part (b), the value for $P\left( {A \cup B} \right)$ is 0.4.

The required probability is,

$\begin{array}{c}\\P\left( {A \cup B} \right)' = 1 - P\left( {A \cup B} \right)\\\\ = 1 - 0.4\\\\ = 0.6\\\end{array}$

(e)

The value for $P\left( {A' \cup B} \right)'$ is obtained as shown below:

From the information, $P\left( A \right) = 0.3\,\,,P\left( {A\, \cap \,B} \right) = 0.1$

The required probability is,

$\begin{array}{c}\\P\left( {A' \cup B} \right)' = 1 - P\left( {A' \cup B} \right)\\\\ = 1 - \left[ {P\left( {A'} \right) + P\left( B \right) - P\left( {A' \cap B} \right)} \right]\\\\ = 1 - \left[ {0.7 + 0.2 - 0.1} \right]\\\\ = 1 - 0.8\\\\ = 0.2\\\end{array}$

Ans: Part a

The probability the complement of event A is 0.7.