# Reviewing THEORETICAL PROBABILITY using PLAYING CARDS (PREVIEW ONLY - some features are disabled) - PowerPoint PPT Presentation

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Reviewing THEORETICAL PROBABILITY using PLAYING CARDS (PREVIEW ONLY - some features are disabled). INTRO. LESSON. QUIZ. Topics in this Lesson. 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).

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

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

INTRO

LESSON

QUIZ

### Topics in this Lesson

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!

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### Probability of an event

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

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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.

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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).

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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.

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### Probability of an event NOT occurring

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

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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.

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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.

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### “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)

EXAMPLE:

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

52 playing cards

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“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.

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“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.

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“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.

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“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.

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### “AND” Probability with two independent events

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?

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### “AND” Probability with two independent events

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.

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### “AND” Probability with two independent events

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.

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### “AND” Probability with two independent events

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.

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### “AND” Probability with two independent events

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

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### “AND” Probability with two Dependent Events

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?

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### “AND” Probability with two Dependent Events

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.

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### “AND” Probability with two Dependent Events

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.

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### “AND” Probability with two Dependent Events

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.

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### “AND” Probability with two Dependent Events

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.

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### “AND” Probability with two Dependent Events

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.

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### CREDITS

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

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