Question

A) Lance the Wizard has been informed that tomorrow there will be a 50% chance of encountering the evil Myrmidons and a 30% chance of meeting up with the dreadful Balrog. Moreover, Hugo the Elf has predicted that there is a 10% chance of encountering both tomorrow. What is the probability that Lance will be lucky tomorrow and encounter neither the Myrmidons nor the Balrog?

B)The astrology software package, Turbo Kismet,† works by first generating random number sequences and then interpreting them numerologically. When I ran it yesterday, it informed me that there was a 1/3 probability that I would meet a tall dark stranger this month, a 2/3 probability that I would travel this month, and a 1/9 probability that I would meet a tall dark stranger and also travel this month. What is the probability that I will either meet a tall dark stranger or that I will travel this month? HINT [See Example 4.] (Enter your probability as a fraction.)

Answer 1

In each of these examples we will clear the clutter of events by using more compact notation.

A)

Denote the events as:

: Encountering the evil Myrmidons B: Meeting up with the dreadful Balrog

What is given to us?

The probability of encountering the evil Myrmidons, i.e.
| PIA0.5
. The probability of meeting up with the dreadful Balrog, i.e.
PB0.3
., and $P[A \cap B]=0.1$ i.e the probability of encountering both of them.

Now we want the probability that Lance doesn't encounter any of the the two. Firstly, this event is $A^\complement \cap B^ \complement=(A \cup B)^ \complement$

Also, $P[(A \cup B)^ \complement]=1-P[A \cup B]$

$P[A \cup B]=P[A]+P[B]-P[A \cap B]=0.5+03-0.1=0.7$

Hence, is the answer.

PUAU B)cj = 1-0.7-0.6

B) Again define the following events:

A: Meeting a tall dark stranger B: Traveling this month

What's given to us?

Probability of meeting a tall dark stranger,
3 P[A] =
. Probability of traveling this month,
P[B] =っ
. Also, the probability of both these things happening, $P[A \cap B]=\dfrac{1}{9}$ .

We want the probability that only one of these events happen. This is either: $A \cap B^\complement$ or $A^\complement \cap B$ .

Let's do it one at a time:

Note that $(A^\complement \cap B) \cup (A \cap B)=B$ . Also, $(A^\complement \cap B) \ \text{and} \ (A \cap B)$ are mutually exclusive.

Therefore,

$\\ P[(A^\complement \cap B) \cup (A \cap B)]=P[B] \\ P[A^ \complement \cap B]+P[A \cap B]=P[B] \\ P[A^ \complement \cap B]+ \dfrac{1}{9}=\dfrac{2}{3} \\ P[A^ \complement \cap B]=\dfrac{2}{3}-\dfrac{1}{9}=\dfrac{5}{9}$

Similarly, $(A \cap B^\complement) \cup (A \cap B)=A$ . Also, $(A\cap B^\complement ) \ \text{and} \ (A \cap B)$ are mutually exclusive.

Therefore,

$\\ P[(A \cap B^\complement) \cup (A \cap B)]=P[A] \\ P[A \cap B^\complement]+P[A \cap B]=P[A] \\ P[A \cap B^\complement]+ \dfrac{1}{9}=\dfrac{1}{3} \\ P[A \cap B^\complement]=\dfrac{1}{3}-\dfrac{1}{9}=\dfrac{2}{9}$

Thus, the required probability is: $P[(A^\complement \cap B) \cup (A \cap B^\complement)]$ . Now note that they're both mutually exclusive.

$P[(A^\complement \cap B) \cup (A \cap B^\complement)]=P[A^\complement \cap B]+P[A \cap B^\complement]=\dfrac{5}{9}+\dfrac{2}{9}=\dfrac{7}{9}$

This is the final answer.