Section 6-4
(8) P(A') = 60/100 = 3/5 = 0.6 = 60%
This could happen only on 2 out of the 6 possible rolls; if you get a 3 or a 6. Thus the probability of this event is P(E) = 2/6. The odds are P(E)/P(E') = 2/4 = 1/2.
The sample space S consists of the 52 cards.
Let A = {K♥, K♦, K♣, K♠} be the event of drawing a
king.
Let B = {A♥, 2♥, 3♥, 4♥, 5♥, 6♥, 7♥,
8♥, 9♥, 10♥, J♥, Q♥, K♥} be the event
of drawing a heart.
Then A ∩ B = {K♥} has just 1 element.
We are interested in the probability of the event E = A ∪ B, namely drawing a King or a heart.
Now, the probability of drawing a King or a heart is
P(E) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 4/52 + 13/52 - 1/52 = 16/52
= 4/13 ~ 30.76923%
The odds of drawing a King or a heart are P(E)/P(E') = (4/13)/(9/13) = 4/9.
(46)
What is the probability of getting at least one black card in a 7-card hand
off a shuffled 52-card deck?
The sample space S is the set of all 7-card hands, so n(S) = C52, 7
The event E we want to calculate the probability of consinsts of all possible
7-card hands which have at least one black card.
It's easier describe the complement of this event, namely the event E' of getting all red cards in the 7-card hand. Now, such a hand is formed by selecting 7 cards from the 26 red cards in the deck, so n(E') = C26, 7 .
By the Complement Rule,
P(E) = 1 - P(E') = 1 - n(E')/n(S) = 1 - ( C26, 7 )/( C52, 7
) ~ 0.99508 ~ 99.5%
(54)
In a group of n ≤ 100 people, each person writes a number between 1 and 100
on a slip of paper, and puts it in a hat. What's the cance that two or more
pieces of paper have the same number on them?
A single (simple) event in the sample space would consist of n pieces of paper (one per person), each with a number between 1 and 100 on it. Thus we may regard an event in S as consisting of n blanks
_ _ _ _ _ ... _ _ _
filled in with numbers between 1 and 100. Therefore, n(S) = 100n.
We want to compute P(E), where E ⊂ S consists of all simple
events (outcomes) that have at least one number repeated. It's easier deal with
the complement E' of E. The event E' consists of all outcomes having no repeated
number, that is outcomes for which it is not the case that two or more
slots have the same number. How many such outcomes are there? Well, the
number of such outcomes equals the number or ways to arrange in the slots n
different numbers between 1 and 100.
Thus n(E') = P100, n = 100! / (100-n)!.
Therefore, using the complement rule,
P(E) = 1 - P(E') = 1 - n(E')/n(S) = 1 - 100! / ( 100n (100 - n)!
).
|
MALE
|
FEMALE
|
TOTAL
|
|
|
RHESUS MONKEYS
|
3
|
7
|
10
|
|
CHIMPS
|
6
|
4
|
10
|
|
DOGS
|
2
|
8
|
10
|
|
TOTAL
|
11
|
19
|
30
|
One animal is selected at random. The sample space is the set of all 30 animals.
What are the chances of getting:
(A) A chimp or a dog?
Let C and D be the events "choosing a dog" and choosing a chimp,"
respectively.
Thus we want the probability of the event C ∪ D.
Now, P( C ∪ D) = P(C) + P(D) - P(C ∩ D) = P(C) + P(D) - P(∅) =
10/30 + 10/30 - 0 = 20/30 = 2/3 ~ 66.6%.
(B) A chimp or a male?
Let C be the event of choosing a chimp. Let M be the set of choosing a male.
Then C ∪ M is the event of choosing a chimp or a male.
P( C ∪ M) = P(C) + P(M) - P(C ∩ M) = 10/30 + 11/30 - 6/30 = 15/30 =
1/2 = 50%
(C) An animal other than a female monkey?
Exactly 30-7 = 23 of the 30 animals are not female monkeys, so the chance of
choosing one of them is 23/30 ~ 76.6%