Exercise 2 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 sum of

10

" and

B

be "doubles." If we roll two

5 s,

the sum is

10

and the roll is doubles. Therefore,

A

and

B

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

P (A),

P (B),

and

P (A and B) .

Think about all possible pairs of numbers from dice which result in rolling a sum of

10

and all possible doubles.

\underline{Sum = 10} 5, 5 4, 6 6, 4 \underline{Doubles} 1, 1 2, 2 3, 3 4, 4 5, 5 6, 6

Because there are two dice and both can result in

6

outcomes, we have

6 \times 6 = 36

possible outcomes.

P (A) P (B) P (A and B) = \frac{3}{3 6} \leftarrow \leftarrow sum of 10 possible outcomes = \frac{6}{3 6} \leftarrow \leftarrow doubles possible outcomes = \frac{1}{3 6} \leftarrow \leftarrow sum of 10 and doubles possible outcomes

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{3}{3 6} + \frac{6}{3 6} - \frac{1}{3 6}

AddSubFrac

Add and subtract fractions

P (A or B) = \frac{8}{3 6}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

P (A or B) = \frac{2}{9}

We found that

P (sum of 10 or doubles) = \frac{2}{9} .

b Let

A

be "a sum of

6

" and

B

be "a sum of

7 .

" A sum of two numbers cannot be

6

and

7

at the same time, so

A

and

B

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

P (A or B)

=

P (A)

+

P (B) .

Think about all possible pairs of numbers from dice which result in rolling a sum of

6

and

7 .

\underline{Sum = 6} 1, 5 5, 1 2, 4 4, 2 3, 3 \underline{Sum = 7} 1, 6 6, 1 2, 5 5, 2 3, 4 4, 3

Because there are two dice and both can result in

6

outcomes, in total we have

6 \times 6 = 36

possible outcomes.

P (A) P (B) = \frac{5}{3 6} \leftarrow \leftarrow sums of 6 possible outcomes = \frac{6}{3 6} \leftarrow \leftarrow sums of 7 possible outcomes

Now let's calculate

P (A or B) .

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

SubstituteValues

Substitute values

P (A or B) = \frac{5}{3 6} + \frac{6}{3 6}

AddFrac

Add fractions

P (A or B) = \frac{1 1}{3 6}

We found that

P (sum of 6 or 7) = \frac{1 1}{3 6} .

c Let

A

be "a sum

< 3

" and

B

be "a sum

> 10 .

" A sum of two numbers cannot be

< 3

and

> 10

at the same time, so

A

and

B

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

P (A or B)

=

P (A)

+

P (B) .

Think about all possible pairs of numbers from dice which result in rolling a sum

< 3

and

> 10 .

\underline{Sum < 3} 1, 1 \underline{Sum > 10} 5, 6 6, 5 6, 6

Because there are two dice and both can result in

6

outcomes, in total we have

6 \times 6 = 36

possible outcomes.

P (A) P (B) = \frac{1}{3 6} \leftarrow \leftarrow sum < 3 possible outcomes = \frac{3}{3 6} \leftarrow \leftarrow sum > 10 possible outcomes

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}{3 6} + \frac{3}{3 6}

AddFrac

Add fractions

P (A or B) = \frac{4}{3 6}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

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

We found that

P (sum < 3 or sum > 10) = \frac{1}{9} .

Subchapter links

Exercises p.15-P15