Question

The table below shows the results of a survey that asked 1060 adults from a certain country if they would support a change in their country's flag. A person is selected at random. Complete parts (a) through (d).

 

(a) Find the probability that the person opposed the change or is female.

P(opposed the change or female) =

(Round to the nearest thousandth as needed.)

(b) Find the probability that the person supports the change or is male.

P( supports the change or male) =

(Round to the nearest thousandth as needed.)

(c) Find the probability that the person is not unsure or is female. P (not unsure or female) =

(Round to the nearest thousandth as needed.)

(d) Are the events "being male" and "support" mutually exclusive? Explain.

A. Yes, because one can't be male and support the change at the same time.

B. No, because there are 155 males that support the change.

C. Yes, because there are 155 males that support the change.

D. No, because one can't be male and support the change at the same time.

Answer 1

Concepts and reason

Probability: The ratio of the number of favorable outcomes to certain event and total number of possible outcomes is called as the probability of an event.

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

Mutually Exclusive events: Two events are said to be mutually exclusive if they cannot occur at the same time. It is also known as disjoint event.

Fundamentals

The probability of an event is defined as,

The probability of not an event is defined as,

$\begin{array}{c}\\P\left( {E'} \right) = 1 - \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}\,{\rm{for}}\,{\rm{an}}\,{\rm{event}}}}{{{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{outcomes}}}}\\\\ = 1 - \frac{{N\left( E \right)}}{{N\left( S \right)}}\\\end{array}$

The probability of events A and B is defined as,

${\rm{P}}\left( {A \cap B} \right) = \frac{{N\left( {A \cap B} \right)}}{{N\left( S \right)}}$

Addition Rule of Probability:

Let A and B be two events, then the probability for event A or event B can be defined as,

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

If the events A and B are mutually exclusive or disjoint then, $P\left( {A \cap B} \right) = 0$

(a)

The probability that the person opposed the change or female is obtained as shown below:

Let S denote the sample space. Define event A as the person who opposes a change in their country’s flag and event B as female.

From information given, $N\left( A \right) = 625$ , $N\left( B \right) = 580$ , $N\left( {A \cap B} \right) = 307$ and $N\left( S \right) = 1,060$ .

The probability is,

$\begin{array}{c}\\P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\\\ = \frac{{N\left( A \right)}}{{N\left( S \right)}} + \frac{{N\left( B \right)}}{{N\left( S \right)}} - \frac{{N\left( {A \cap B} \right)}}{{N\left( S \right)}}\\\\ = \frac{{625}}{{1,060}} + \frac{{580}}{{1,060}} - \frac{{307}}{{1,060}}\\\\ = \frac{{898}}{{1,060}}\\\\ = 0.8472\\\end{array}$

(b)

The probability that the person supports the change or male is obtained as shown below:

Define the event C as the person who supports the change in their country’s flag and event D as male.

From information given, $N\left( C \right) = 400$ , $N\left( D \right) = 480$ , $N\left( {C \cap D} \right) = 155$ and $N\left( S \right) = 1,060$

The probability is,

$\begin{array}{c}\\P\left( {C \cup D} \right) = P\left( C \right) + P\left( D \right) - P\left( {C \cap D} \right)\\\\ = \frac{{N\left( C \right)}}{{N\left( S \right)}} + \frac{{N\left( D \right)}}{{N\left( S \right)}} - \frac{{N\left( {C \cap D} \right)}}{{N\left( S \right)}}\\\\ = \frac{{400}}{{1,060}} + \frac{{480}}{{1,060}} - \frac{{155}}{{1,060}}\\\\ = \frac{{725}}{{1,060}}\\\\ = 0.6840\\\end{array}$

(c)

Define event E as the person is unsure.

Consider, $N\left( S \right) = 1,060$ represents the total number of persons. Also, the number of unsure persons, $N\left( E \right) = 38$ .

The probability is,

$\begin{array}{c}\\P\left( {E'} \right) = 1 - \frac{{N\left( E \right)}}{{N\left( S \right)}}\\\\ = 1 - \frac{{38}}{{1,060}}\\\\ = 1 - 0.0358\\\\ = 0.9642\\\end{array}$

Define $E' \cap B$ be the event that the person is not unsure and female.

Consider, $N\left( S \right) = 1,060$ represents the total number of persons. Also, the number of not unsure and female,

$\begin{array}{c}\\N\left( {E' \cap B} \right) = 245 + 307\\\\ = 552\\\end{array}$ .

The probability is,

$\begin{array}{c}\\P\left( {E' \cap B} \right) = \frac{{N\left( {E' \cap B} \right)}}{{N\left( S \right)}}\\\\ = \frac{{552}}{{1,060}}\\\\ = 0.5208\\\end{array}$

The probability that the person is not unsure or is female is obtained as shown below:

Define the event E’ is not unsure and event B denotes female.

From information given, $N\left( {E'} \right) = 1,022$ , $N\left( B \right) = 580$ , $N\left( {E' \cap B} \right) = 552$ and

$N\left( S \right) = 1,060$ .

The required probability is,

$\begin{array}{c}\\P\left( {E' \cup B} \right) = P\left( {E'} \right) + P\left( B \right) - P\left( {E' \cap B} \right)\\\\ = \frac{{N\left( {E'} \right)}}{{N\left( S \right)}} + \frac{{N\left( B \right)}}{{N\left( S \right)}} - \frac{{N\left( {E' \cap B} \right)}}{{N\left( S \right)}}\\\\ = \frac{{1,022}}{{1,060}} + \frac{{580}}{{1,060}} - \frac{{552}}{{1,060}}\\\\ = \frac{{1,050}}{{1,060}}\\\\ = 0.9906\\\end{array}$

(d)

The events being male and support mutually exclusive is obtained as shown below:

The event C as the person who supports the change in their country’s flag and event D denote male.

From information given, $N\left( {C \cap D} \right) = 155$ and $N\left( S \right) = 1,060$ .

The probability is,

$\begin{array}{c}\\P\left( {D \cap C} \right) = \frac{{N\left( {D \cap C} \right)}}{{N\left( S \right)}}\\\\ = \frac{{155}}{{1,060}}\\\\ = 0.1462\\\\ \ne 0\\\end{array}$

Thus, the probability of being male and support is not equal to zero. This implies that, the events are not mutually exclusive.

Ans: Part a

The probability that the person opposed the change or female is 0.8472.