Transcript Independence/Multiplication Rule

Probability, Part III
Learning Objectives
By the end of this lecture, you should be able to:
– Describe what is meant by independent and non-independent
events.
– Describe and apply the multiplication rule for independent
events.
– Describe two common ways of coming up with probabilities.
– Compare and contrast the concepts of disjointness v.s.
independence.
Methods of determining the probability to an event:
There are 2 well-established ways of assigning probabilities with some degree of
accuracy:
• Theoretically  from our understanding of the phenomenon and symmetries in the
problem
– For a 6-sided fair die, P(2) = 1/6
– For a deck of cards, P(Ace of Hearts) = 1/52
• Empirically  We look at data. In other words, we are deriving our knowledge based
on numerous similar past events. We determine a probability empirically when there
isn’t a theoretical method of determining the probability.
– EXAMPLE: What is the probability that a random student in IT-223 wil receive an A? This
probability can not be determine theoretically the way, say, a coin flip can be. Instead, we’d
have to look at past data from IT-223 courses and see how many As were recorded. If in a
sample of 600 students, 54 A’s were recorded, then P(grade of A) = 54 / 600.
– EXAMPLE: What is the probability of a baseball player getting a hit on a given at bat? Again,
we would have to look at past data to see his/her batting average. If over the last year, he
hit 122 times out of 411 at bats, P(Hit) = 122/400.
An easy way of determining a probability when all outcomes are equally likely:
When all outcomes are equally likely (have the same probability), there is a nice
quick way of calculating probabilities:
If all of the outcomes are equally likely (i.e. they all have the same
probabilities), such as a die roll or a coin flip, then each individual outcome has
probability: 1 / (# of possible outcomes).
•
Eg: For a coin flip, all possible outcomes have an equal probability of 1 / (2 possible
outcomes) = 0.5
•
Eg: For a die roll, all outcomes have an equal probability of 1 / (6 possible outcomes)
= 0.167
Formula: For any event A where all outcomes have the same probability:
P(A) 
count of outcomes in A
count of outcomes in S
Dice
You toss two dice. What is the probability of the outcomes summing to 5?
S:
{(1,1), (1,2), (1,3),
……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice).
Thus, the probability of any one of them is 1/36.
P(sum of 5) = 4 possible outcomes
P(outcomes in Sample space) = 36 outcomes in sample space
= 4/36
EXAMPLE: What is the probability of the ball landing in an even number slot?
Answer: All slots have an equal probability. So:
P(Even)
= count of Even outcomes / count of All outcomes
= 18 / 38
= 0.47
count of outcomes in A
P(A) 
count of outcomes in S
Probability rules (cont’d)
Multiplication Rule for Independent Events
5) Two events A and B are independent if knowing that one event has occured
does not change the probability that the other will occur.
Independence
•
Flip a coin once and get ‘heads’. Does this result change the probability of rolling a
heads again on the second roll?
– Event 1: Get a heads on first flip. Event 2: Get a heads on second flip. The probability
of event #2 is NOT affected by whatever happened on event #1. Therefore, these are
independent events
•
You buy a lottery ticket and win. Then you win again the next week. Does this
change the likelihood of winning on the third week since you are “on a roll”?
– Event 1: Win the first week. Event 2: Win the second week. Event 3: Win the third
week. However, winning the first two weeks in no way affects the likelihood of winning
the 3rd week. In other words, these events are all independent
•
In a game of Poker, you draw a card from the deck and get an Ace. Does this result
change the probability of getting an Ace on a second draw from that same deck?
– Event 1: First card is an Ace. Event 2: Second card is an Ace. The probability of event #2
IS affected by whatever happened in event #1. Therefore, these events are NOT
independent. (Initially, 4/52 chance, after removing 1 ace, 3/51 chance)
Two events are independent if the probability that one event occurs on
one trial of an experiment is not affected or changed by the occurrence
of any other event (e.g. two consecutive coin tosses) .
When are events NOT independent?
Imagine that these coins were spread out so that half the coins showed heads and
half showed tails. Close your eyes and pick one. The probability of it being heads
is 0.5. However, if you don’t put it back in the pile, the probability of picking up
another coin that is heads up is now less than 0.5.
The trials could only be considered
independent if you put the coin back
each time.
Probability rules (cont’d)
Multiplication Rule for Independent Events
5) Two events A and B are independent if knowing that one event has occured
does not change the probability that the other will occur.
This rule leads us to the multiplication rule for independent events:
P(A and B) = P(A) * P(B)
Again, this rule only applies to independent events. Later we will show a second rule that applies to
non-independent events
“AND” Questions
• Recall how in a previous lecture, we discussed “or” questions.
– E.g. I need to decide what to wear to work: What is the probability that it will be snowing
or it will be raining?
• Well, we also frequently encounter “AND” questions:
– I need to roll a 5 and a 2 in a game of backgammon to win the game.
– I need to draw two cards in which one must be an Ace and the other
must be a 7.
For “and” questions, we use the multiplication rule.
Example
Multiplication Rule for Independent Events
Example: What is the probability of getting a tails on two consecutive coin tosses?
We are asking the probability of getting a tails on the first coin toss and on the second
coin toss. With ‘and’ questions, we typically think of the multiplication rule. At that point
we must ask ourselves if the events are independent.
Because we are discussing two consecutive coin tosses, which means that whatever
happens in the first coin toss does not affect what happens on the second coin toss, the
events are indeed indepdendent. Therefore, we may use our multiplication ruel for
independent events:
P(first = Tail and second = Tail)
= P(first Tail) * P(second Tail)
= 0.5 * 0.5
= 0.25
Example:
A couple intends to have three children. What is the likelihood that they will have only
boys?
Answer: After a little bit of thought, you realize that you are being asked P(first child
is a boy) AND P(2nd child is a boy) AND P(3rd child is a boy). ‘AND’ questions should
make you think of the multiplication rule. At that point, you need to ask yourself if
these 3 events are independent. In fact, they are. So we may therefore use the
multiplication rule:
P(BBB) = P(B)* P(B)* P(B) = (1/2)*(1/2)*(1/2) = 1/8
Example
• You are dealt two cards out of a deck. Calculate the probability the the
cards are the Ace of Spades and the Ace of Hearts.
– Solution: You will probably be tempted to use the multiplication rule.
However, this rule (in its current form) only applies to independent events. As
it turns out, these events are not independent. If one of the events has
occurred (e.g. drawing the ace of spades from the deck), the probability of the
second event changes as a result.
• Later, we will learn a slight modification to this rule so that it can also be applied to
non-independent events.
– If, however, we changed the question to state that instead of pulling two cards
out of the deck, we pull out one card, then put it back and pull out another
card, we could use our multiplication rule, since these are now independent
events.
Example (from text)
•
A state lottery’s Pick 3 game asks players to choose a 3-digit number, 000 to 999.
The state chooses the winning number at random, so each number has probability
1/1000. You win if the winning number contains the digits in your number, in any
order.
– Your number is 456. What is your probability of winning? (Write on a piece of paper)
– Your number is 212. What is your probability of winning.
•
Solution: As always, you have to read the question carefully, and always be sure to
watch the fine print! In this case, the three tiny words at the end, “in any order”
change the entire nature of the problem.
– So, instead of winning with 456 where your probability would be 1/1000, you
can also win with 465, or 546, or 564 etc. If you look at all possible
permutations, you will see that there are 6 distinct possibilities: {456, 465,
546, 564, 645, 654}. Since each one has a 1/1000 chance of winning, and you
have six possibilities, the probability of winning is 6/1000 or 0.006.
– For the second part, you only have three distinct arrangements: {212, 221,
122}. So your probability of winning is 1/1000 + 1/1000 + 1/1000 = 0.003.
Incidentally, did you try to use the multiplication rule here? In fact, you should be using the addition
rule! Remember that you must always make sure you are using the right tool for the job!!
Example (from text)
•
The PINs for ATMs usually consist of 4 digits. You notice that most PINs have at least
one 0 and you wonder if the issuers use lots of 0s to make the numbers easy to
remember. Suppose that PINs are assigned at random so that all 4-digit numbers are
equally likely.
– How many possible PINs are there? – Write down your answer
– What is the probability that a PIN assigned at random has at least one 0? – Write.
SOLUTION:
• Possible PINs: 104 = 10,000
• The operative word for the second question is “at least”. In other words, what is the
probability that 1 or 2 or 3 or all 4 numbers are zero? This turns out to be a
surprisingly involved calculation. However, if you remember your trusty complement
rule, it becomes surprisingly easy! You realize that you could phrase things another
way: If I can determine the probability that NONE of the numbers are 0, then the complement of that
equals the probability at AT LEAST one is 0.
• P(one number = 0) = 0.1, P(not 0)=0.9
• P(all 4 numbers are not 0) = 0.9*0.9*0.9*0.9 = 0.6561. Don’t forget that this is the
probability that NONE are 0. So the probability that 1 or more are zero = 1-0.06561 =
0.3439.
Disjoint vs Independent
• The two are often confused. Be sure that you are clear on the meaning of
each.
• One common question has to do with the relationship between them. Is
there one? For example, can an event be both disjoint and independent?
1. Disjoint = If two events, A and B are disjoint, we are saying that if the event ‘A’
A is true, then P(B) can not be true. So let’s say that our event is indeed
disjoint.
2. If our event is disjoint, this says that P(B) is affected by A. Well, by definition, if
the probability of B (i.e. P(B) ) is indeed changed depending on whether or
not A occurs , then P(B) is indeed DE-pendent on A!
3. So the answer to our question is NO! When you think about it, if two events
are disjoint, then those events can NOT be independent.
I fully recognize that you may need to spend some time wrapping your brain
around this. That’s okay – however, it is something that you should do. Think of it
as ‘brain exercise’! (Plus it helps to make sure you understand the concept).