- 62 Views
- Uploaded on
- Presentation posted in: General

Reviewing THEORETICAL PROBABILITY using PLAYING CARDS (PREVIEW ONLY - some features are disabled)

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Reviewing THEORETICAL PROBABILITYusingPLAYING CARDS(PREVIEW ONLY - some features are disabled)

INTRO

LESSON

QUIZ

1. The probability of an event, E, is:

P(E) =

4. The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

number of ways E can occur

total number of outcomes

P(A and B) = P(A) X P(B given that A has occurred)

2. The probability of an event, E, NOT occurring is:

P(not E) = 1 – P(E)

5. The probability of dependent events

A and B occurring is:

P(A and B) = P(A) X P(B|A)

3. The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

By clicking the “NEXT” button, you will begin the lesson with topic 1. If you wish

to go directly to a topic, click on that topic. At the end of the lesson is a brief quiz

covering all 5 topics. Good luck and enjoy this review of probability!

HOME

NEXT

LAST

The probability of an event, E, is:

P(E) =

number of ways E can occur

total number of outcomes

EXAMPLE:

Let’s consider the event of drawing a king from a deck of 52 playing cards

HOME

NEXT

LAST

Probability of an event

The probability of an event, E, is:

P(E) =

number of ways E can occur

total number of outcomes

Because there are 4 kings in a deck of cards (1 of each suit), then 4 is the number of ways E can occur.

HOME

PREV

NEXT

LAST

Probability of an event

The probability of an event, E, is:

P(E) =

number of ways E can occur

total number of outcomes

Because there are 52 cards, then 52 is the total number of outcomes(of drawing a card).

HOME

PREV

NEXT

LAST

Probability of an event

The probability of an event, E, is:

P(E) =

number of ways E can occur

total number of outcomes

Therefore, the probability of drawing a king from a deck of cards is 4/52. If we reduce the fraction, then the probability is 1/13.

HOME

PREV

NEXT

LAST

The probability of an event, E, NOT occurring is:

P(not E) = 1 – P(E)

EXAMPLE:

Let’s consider the event of drawing a card that is NOT a king from a deck of 52 playing cards

HOME

PREV

NEXT

LAST

Probability of an event NOT occurring

The probability of an event, E, not occurring is:

P(not E) = 1 – P(E)

Remember, P(E) is the probability of drawing a king from a deck of cards and we calculated that value to be 4/52.

HOME

PREV

NEXT

LAST

Probability of an event NOT occurring

The probability of an event, E, not occurring is:

P(not E) = 1 – P(E)

So the probability of drawing a card that is NOT a king from a deck of cards is 1 – 4/52 or 48/52. If we reduce the fraction, then the probability is 12/13.

HOME

PREV

NEXT

LAST

.

The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

EXAMPLE:

Let’s consider the event of drawing a heart or a king from a deck of

52 playing cards

HOME

PREV

NEXT

LAST

“OR” Probability of two events

.

The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

There are 13 hearts (out of 52 total cards) in a deck of cards. That’s event A and P(A) = 13/52.

HOME

PREV

NEXT

LAST

“OR” Probability of two events

.

The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

Then let’s look at event B which is drawing a king. There are 4 kings (out of 52 total cards). So, P(B) = 4/52.

HOME

PREV

NEXT

LAST

“OR” Probability of two events

.

The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

Finally, there is 1 card that is both a king and a heart. That’s the event of both A and B occurring. So P(A and B) = 1/52.

HOME

PREV

NEXT

LAST

“OR” Probability of two events

.

The probability of event A or event B occurring is:

P(A or B) = P(A) + P(B) – P(A and B)

Now, let’s put it all together. P(A) = 13/52. P(B) = 4/52. P(A and B) = 1/52. Therefore, P(A) + P(B) – P(A and B) = 13/52 + 4/52 – 1/52 = 16/52.

HOME

PREV

NEXT

LAST

The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

EXAMPLE:

Suppose we have two decks of cards. Let’s consider the event of drawing one card

from each deck. What is the probability of drawing an ace of spades and a red 10?

HOME

PREV

NEXT

LAST

The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

In this case, drawing an ace of spades and a red 10 are independent events because

we’re drawing each card from two separate decks.

The outcome of one event has no effect on the outcome of the other event.

HOME

PREV

NEXT

LAST

The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

The probability of drawing an ace of spades from a deck of 52 cards is 1/52.

So P(A) = 1/52.

HOME

PREV

NEXT

LAST

The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

There are 2 red 10’s in a deck of cards; the 10 of diamonds and the 10 of hearts.

So, the probability of drawing a red 10 from a deck of 52

cards is 2/52. Reduced, that means P(B) = 1/26.

HOME

PREV

NEXT

LAST

The probability of independent events A and B occurring is:

P(A and B) = P(A) X P(B)

Finally, since P(A) = 1/52 and P(B) = 1/26, then the probability of

drawing an ace of spades from one deck and a red 10 from another deck is:

P(A) X P(B) = 1/52 X 1/26 = 1/1352

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

EXAMPLE:

Let’s consider the event of drawing two cards from the one deck.

Now what is the probability of drawing an ace of spades and a red 10?

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

In this case, drawing an ace of spades and a red 10 are dependent events because

we’re drawing the two cards from the same deck.

The outcome of one event has an effect on the outcome of the other event.

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

The probability of drawing an ace of spades from a deck of 52 cards is 1/52.

So P(A) = 1/52.

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

With the ace of spades drawn, we now have only 51 cards remaining from which to draw a red 10.

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

The probability of drawing a red 10 from 51 cards is 2/51.

So, P(B|A) = 2/51.

HOME

PREV

NEXT

LAST

The probability of dependent events A and B occurring is:

P(A and B) = P(A) X P(B given that A has occurred), written

P(A and B) = P(A) X P(B|A)

Therefore, given dependent events A and B, the probability of drawing an ace of spades and a red 10 from a deck of cards is 1/52 X 2/51 = 2/2652 = 1/1326.

HOME

PREV

NEXT

LAST

The following sources deserve credit in part for this non-linear PowerPoint

presentation.

“Thinking Mathematically” by Robert Blitzer, Prentice Hall

Microsoft Office Online (office.microsoft.com)

jfitz.com

flickr.com

HOME