Exercise 4 Page P15

a Events that cannot occur at the same time are mutually exclusive. Mutually exclusive events have no outcomes in common. For example, because it is not possible to toss a coin and obtain heads and tails at the same time, these two events are mutually exclusive.

Addition Rules for Probability
If $A$ and $B$ are mutually exclusive events, the probability that $A$ or $B$ will occur is $P (A or B) = P (A) + P (B) .$	If $A$ and $B$ are not mutually exclusive events, the probability that $A$ or $B$ will occur is $P (A or B) = P (A) + P (B) - P (A and B) .$

Let

A

be a student is a freshman and

B

be a student is female. Nine of the ten freshmen are female, so it is possible to be both female and a freshman in the given French class. Therefore,

A

and

B

are not mutually exclusive events. Let's start by calculating

P (A),

P (B),

and

P (A and B) .

P (A) P (B) P (A and B) = \frac{1 0}{2 0} \leftarrow \leftarrow freshmen total students = \frac{1 2}{2 0} \leftarrow \leftarrow females total students = \frac{9}{2 0} \leftarrow \leftarrow female freshmen total students

Finally, considering that

A

and

B

are not mutually exclusive events, let's calculate

P (A or B) .

P (A or B) = P (A) + P (B) - P (A and B)

SubstituteValues

Substitute values

P (A or B) = \frac{1 0}{2 0} + \frac{1 2}{2 0} - \frac{9}{2 0}

AddSubFrac

Add and subtract fractions

P (A or B) = \frac{1 3}{2 0}

We found that

P (freshman or female) = \frac{1 3}{2 0} .

b Let

A

be a student is a sophomore and

B

be a student is male. In this exercise, we are given that

2

of the

8

sophomores are female. Thus,

6

sophomores are male and

A

and

B

are not mutually exclusive events. Remember, for events that are not mutually exclusive,

P (A or B)

=

P (A)

+

P (B)

-

P (A and B) .

P (A) P (B) P (A and B) = \frac{8}{2 0} \leftarrow \leftarrow sophomores total students = \frac{8}{2 0} \leftarrow \leftarrow males total students = \frac{6}{2 0} \leftarrow \leftarrow male sophomores total students

Considering that

A

and

B

are not mutually exclusive events, let's calculate

P (A or B) .

P (A or B) = P (A) + P (B) - P (A and B)

SubstituteValues

Substitute values

P (A or B) = \frac{8}{2 0} + \frac{8}{2 0} - \frac{6}{2 0}

AddSubFrac

Add and subtract fractions

P (A or B) = \frac{1 0}{2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

P (A or B) = \frac{1}{2}

We found that

P (sophomore or male) = \frac{1}{2} .

c Let

A

be a student is a freshman and

B

be a student is a sophomore. A student cannot be a freshman and a sophomore at the same time. Thus,

A

and

B

are mutually exclusive events. Remember, for events that are mutually exclusive,

P (A or B)

=

P (A)

+

P (B) .

P (A) P (B) = \frac{1 0}{2 0} \leftarrow \leftarrow freshmen total students = \frac{8}{2 0} \leftarrow \leftarrow sophomores total students

Now let's calculate

P (A or B) .

P (A or B) = P (A) + P (B)

SubstituteValues

Substitute values

P (A or B) = \frac{1 0}{2 0} + \frac{8}{2 0}

AddFrac

Add fractions

P (A or B) = \frac{1 8}{2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

P (A or B) = \frac{9}{1 0}

We found that

P (freshman or sophomore) = \frac{9}{1 0} .

Subchapter links

Exercises p.15-P15