Data Interpretation & Probability

Finite Basic Probabilities

The probability that an event will occur is a number between zero and one.

The probability of an event that never occurs is zero.
The probability that pigs will learn to fly is zero.

The probability of an event that always occurs is one.
The probability the sun will rise tomorrow is one.

Sample Spaces and Events

A sample space is a set containing ALL possible results for an experiment.

Find the sample space of a coin flipped three times.
Let HHH stand for heads on the first flip, heads on the second flip and heads on the third flip. Let HHT stand for heads on the first flip, heads on the second flip and tails on the third flip. Then the sample space, which is all possible outcomes when flipping three coins, is as follows:
S={HHH, HHT, HTT, HTH, TTT, THH, TTH, THT}

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Definition of Probability

Let S be a finite sample space consisting of equally likely outcomes. Then for any event , the probability of E is given by:

In other words, to find the probability of an event:

Example 1:

Find the probability of flipping a coin three times and getting exactly two tails.

Solution:

Consider the sample space of flipping a coin three times.

S = {HHH, HHT, HTT, HTH, TTT, THH, TTH, THT}

There are 8 possible outcomes for the experiment of flipping a coin three times, so |S| = 8.
The value of |E| is the number of ways you can get what you want. We want exactly two tails, so the possible situations that will satisfy the criteria of exactly two tails is:
|E|={HTT,TTH, THT}
There are 3 ways to satisfy the required two tails, so |E| = 3.

Therefore the probability of flipping a coin three times and obtaining exactly two tails is 3/8.

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Example 2:

During practice, a basketball player is going to shoot three shots.

Find the following:

A)    Find the sample space for this experiment.

Let H be a hit the basketball hoop and M be a miss the basketball hoop.

The possibilities are: Hit, Hit and Hit or Hit, Miss and Miss, etc.

So the sample space is:
S = {HHH, HMM, HMH, HHM, MMM, MHH, MHM, MMH}

B)    Find the sample space for the player misses the third shot.
Consider the above sample space:
S={HHH, HMM, HMH, HHM, MMM, MHH, MHM, MMH}

Find all the possibilities when the player misses the third shot.
E={HMM, HHM, MMM, MHM}

C)    Find the probability the player misses the third shot.
To find the probability, we need |E| and |S|.
The total number of ways the experiment can occur is 8, so |S| = 8.
The total number of ways to have the player miss the third shot is 4, so |E| = 4.

Therefore the probability that the player misses the third shot is .5

D)    Find the sample space for the player makes two shots and then misses the next.
Consider the above sample space:
S = {HHH, HMM, HMH, HHM, MMM, MHH, MHM, MMH}

Find all the possibilities when the player makes two shots and then misses the next.
E={HHM}

E)    Find the probability the player makes two shots then misses.
To find the probability, we need |E| and |S|.
The total number of ways the experiment can occur is 8, so |S| = 8.
The total number of ways to have the player make two shots and then miss the next is 1, so |E| = 1.

Therefore the probability that the player makes two shots then misses is 1/8.
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Example 3:
There were 250,000 tickets sold in a particular lottery.    The following are the number of winning tickets sold for each prize.

#of tickets sold Prize
1 $100,000
10 $10,000
50 $2000
100 $1000

If the winning tickets are selected at random, what is the probability that a random ticket will win the following:

A)    $100,000?
Solution:
The entire sample space is 250,000.
There is only 1 way to get the winning ticket of $100,000.

Therefore the probability is:
^{P($100,000) =}

B)    $2000 or More?
Solution:
The entire sample space is 250,000.
To obtain $2000 or more, a person can win $2000, $10,000 or $100,000.

Therefore the probability is:

P($2000 or More) = P($2000) + P($10,000) + P($100,000)



C)    $2000 or Less?
Solution:
The entire sample space is 250,000.
To obtain $2000 or less, a person can win $2000 or $1000.

Therefore the probability is:

P($2000 or Less) = P($2000) + P($1000)



D)    Nothing?
Solution:
The entire sample space is 250,000.
Since are 161 tickets out of the 250,000 that will win a prize, there are 249,839 tickets that will NOT win a prize.
Therefore the probability is 249,839 divided by 250,000.

Another way to work this problem is with the compliment. As you saw above the probability of winning nothing is all the tickets that do not win divided by all possible tickets. The probability of 1 is all the winning tickets plus all the non-winning tickets. So the probability of winning nothing is:
^{P(No prize) = 1 - P(All Prizes) =}

It seems that if you play this game you are most likely not to win.

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Example 4:

Two fair dice are rolled.
Find the sample space for the experiment.
Solution:
When rolling two dice, one dice can be a 1, and the other dice can be a 1,
or one dice can be a 1, and the other dice can be a 2, etc.

The entire sample space is given by:
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

The number of items in the entire sample space is 36, so    |S| = 36.

If two fair dice are rolled, find the following probabilities:
A)    The sum is seven?
Solution:
There are 36 possible outcomes for the experiment, so |S| = 36.

If a 1 and a 6 is rolled, the sum is seven.
If a 2 and a 5 is rolled, the sum is seven.

E = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}
There are 6 ways to obtain a sum of seven, so |E| = 6.

^{P(Sum is seven)}

Therefore the probability of getting the sum of seven when rolling two dice is 1/6.

B)    Both dice are odd?
Solution:
There are 36 possible outcomes for the experiment, so |S| = 36.

If a 1 and a 1 is rolled, then both dice are odd.
If a 1 and a 3 is rolled, then both dice are odd.

E = {(1,1)(1,3)(1,5)(3,1)(3,3)(3,5)(5,1)(5,3)(5,5)
There are 9 ways to have both dice odd, so |E| = 9.

^{P(Both dice are odd) = 9/36}
= 1/4

Therefore the probability of rolling two dice, and getting both dice odd is 1/4.

C)    The sum is six or both dice are even?
Solution:
There are 36 possible outcomes for the experiment, so |S| = 36.

The event the sum is six is:
{(1,5)(2,4)(3,3)(4,2)(5,1)}

The event that both dice are even is:
{(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(6,2)(6,4)(6,6)}
The roll of (2,4) & (4,2) are in both sets, and should only be counted once!

The possible outcomes for the sum of six or both dice are even is:
E={(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(6,2)(6,4)(6,6)(1,5)(3,3)(5,1)}
There are 12 ways to obtain a sum of six or both dice even, so |E| = 12.
P(The sum is six or both dice are even) = 12/36 = 1/3

Therefore the probability is 1/3.

Student #1:
To win the second prize in a lottery, a player must choose six numbers and correctly match exactly five of the six numbers chosen at random. The choice of numbers are 1 - 44. What is the probability of winning the second prize in this lottery on a single play?

.10 .002 .00003 .30

Answer

Example 5:

Find the probability of winning the California lottery. The possible numbers are numbers 1 - 51. Six numbers are selected and all six must match to win the grand prize.

Solution:
To win the California lottery you must pick 6 numbers, and these 6 numbers must all be the winning numbers. There are 51 numbers in which to make your selections.
Since the order of the winning numbers does not matter, the lottery is a combination.    If the lottery were a permutation, you would not only have to have the winning numbers but they would have to be in the correct order in which the winning numbers were selected.

There are 51 numbers possible, and we are selecting 6, so |S| = C(51,6).

There are 6 winning numbers, and we want all 6, so |E| = C(6,6).

The probability is:

This number is pretty small!
The number .00000006 is very close to zero.

First, lets think about this. What is the probability of winning the lottery if you do not play? Of course the probability is zero. How small does a number need to be before you consider it to be zero?

How about .00001? Is this small enough for you to consider it just about zero? If not, how about .0000001? Is this small enough for you to consider it just about zero?

The probability of winning the California state lottery is .00000006, which is just about zero.
The probability of winning the lottery if you don't play is zero.

It appears that the probability of winning the lottery is pretty much the same as the probability of winning the lottery if you don't play.

How sad for us all!
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Example 6:

Using a standard deck of playing cards, find the probability of drawing 1 king, 2 Aces and 3 Jacks.

Solution:

The number of possible outcomes when drawing 6 cards from a deck of 52 cards is C(51,6), so |S| = C(51,6).

There are 4 Kings in a deck and we want 1: C(4,1)
There are 4 Aces in a deck and we want 2: C(4,2)
There are 4 Jacks in a deck and we want 3: C(4,3)

Since we want 1 King AND 2 Aces AND 3 Jacks, use the Multiplication Principal.

^{P(E) =}

Therefore the probability is .000005.

Student #2:
Using a standard deck of playing cards, find the probability of drawing a full house.

.0014 .05 .0012 .0018

Answer

ODDS
The Statement that the odds are 4 to 3 in favor of event E means the following.    In terms of relative frequency, over the long run E will occur 4 times for every 3 times that it does not occur.

ODDS

If the odds in favor of E are M to N, then:

^{P(E) =}

If    ^{P(E) =}    then the odds are 4 to 3 in favor of E.
The odds against are 3 to 4.

If    ^{P(E) =}    then the odds are 5 to 4 in favor of E.
The odds against are 4 to 5.

Student #3:
Given   ^{P(E) =} .    Find the odds in favor of E and against E.

4 to 9 in favor of E and 9 to 4 against
5 to 4 in favor of E and 4 to 5 against
4 to 5 in favor of E and 5 to 4 against
9 to 4 in favor of E and 4 to 9 against

Answer

Student #4:
Given    P(E) = .85.     Find the odds in favor of E and against E.

17 to 20 in favor of E and 20 to 17 against
17 to 3 in favor of E and 3 to 17 against
3 to 17 in favor of E and 17 to 3 against
20 to 17 in favor of E and 17 to 20 against

Answer

Student #5:
Given the odds in favor of Event E are 9 to 11, find the probability that the event will occur.

.11 .62 .36 .45

Answer

Example 7:

Find the odds in favor of rolling a multiple of 3 when rolling one dice.

Solution:
The total possible outcomes when rolling one dice is 6, so |S| = 6.

To roll a multiple of three, either the number 3 or the number 6 must be rolled.

The probability of rolling either a 3 or a 6 is as follows:

^{P(3 or 6) = P(3) + P(6) =}

^{P(multiple of 3) =}

The odds are 1 to 2 in favor of E.

Therefore the odds are 1 to 2 in favor of E, or 2 to 1 against.

Homework Problems

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Use the subject D2.3

Student #1:    The answer is C:

Solution:
The sample space is all possible outcomes of picking 6 numbers out of 44.
Since the order of the numbers the person picks does not matter, a combination instead of a permutation is used.

|S| = C(44,6) = 7,059,052

We must pick 6 numbers out of 44 and match 5 from the winning numbers. In other words 5 will come from the winning pool of 6 numbers, and the other one will come from the non-winning pool of 38 numbers.

There are 6 winning numbers and we want 5 of them AND there are 38 non-winning numbers and we want 1 of them.

Remember the word AND means to multiply.

|E| = C(6,5) * C(38,1) = 228

^{P(E) =}

Therefore the probability is .00003.

Go Back to Lesson

Student #2:    The answer is A:

Solution:
A full house is 3 of a kind and 2 of a kind.
There are 13 different types of three of a kind:
Three aces, three kings, three queens, ... and three twos.

The number of ways to get 3 aces is: C(4,3).
The number of ways to get 3 kings is: C(4,3).
The number of ways to get 3 queens is: C(4,3).
.
.
.
The number of ways to get 3 twos is: C(4,3).

So there are 13C(4,3) different ways to get any three of a kind.

The other two cards must be two of a kind.
There are 13 different ways to get two of a kind. But we have already used one of the choices when we selected the three of a kind. In other words, if you have three kings then the two of a kind can not be kings.

Thus, the number of ways to get the remaining two of a kind is 12C(4,2).

^{P(full house) =}

Therefore the probability of drawing five cards from a standard deck and obtaining a full house is .0014.

Go Back to Lesson

Student #3:    The answer is C:

Solution:

The two numbers 4 and 5 must add to the total number in the sample space which is 9. Thus, the odds in favor of E are 4 to 5.

The odds are 4 to 5 in favor of E, and 5 to 4 against E.

Go Back to Lesson

Student #4:    The answer is B:

Solution:

^{P(E) = .85 =}

The sum of 17 and 3 equals 20, so the odds are 17 to 3 in favor of E.

The odds in favor of E are 17 to 3, and 3 to 17 against.

Go Back to Lesson

Student #5:    The answer is D:

Solution:

Since the odds are 9 to 11, the total sample space is 20.

^P(E)=

Therefore P(E) is .45.

Go Back to Lesson