reviewing theoretical probability using playing cards (preview only - some features are disabled)

Reviewing THEORETICAL PROBABILITY

usingPLAYING CARDS

(PREVIEW ONLY - some features are disabled)

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Topics in this Lesson1. The probability of an event, E, is:P(E) = number of ways E can occur

total number of outcomes

2. The probability of an event, E, NOT occurring is:P(not E) = 1 – P(E)

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3. The probability of event A or event B occurring is:P(A or B) = P(A) + P(B) – P(A and B)

4. The probability of independent events A and B occurring is:P(A and B) = P(A) X P(B)

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

5. The probability of dependent events A and B occurring is:P(A and B) = P(A) X P(B|A)

By clicking the “NEXT” button, you will begin the lesson with topic 1. If you wishto go directly to a topic, click on that topic. At the end of the lesson is a brief quizcovering all 5 topics. Good luck and enjoy this review of probability!

Probability of an event

The probability of an event, E, is:P(E) = number of ways E can occur


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

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




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


The probability of an event, E, not occurring is:P(not E) = 1 – P(E)

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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.


The probability of an event, E, not occurring is:P(not E) = 1 – P(E)

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

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In this case, drawing an ace of spades and a red 10 are independent events becausewe’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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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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cards is 2/52. Reduced, that means P(B) = 1/26.

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

Finally, since P(A) = 1/52 and P(B) = 1/26, then the probability ofdrawing an ace of spades from one deck and a red 10 from another deck is:



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

“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), writtenP(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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In this case, drawing an ace of spades and a red 10 are dependent events becausewe’re drawing the two cards from the same deck.The outcome of one event has an effect on the outcome of the other event.



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

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The probability of drawing a red 10 from 51 cards is 2/51.So, P(B|A) = 2/51.



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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.

CREDITS

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

“Thinking Mathematically” by Robert Blitzer, Prentice HallMicrosoft Office Online (office.microsoft.com)jfitz.comflickr.com

reviewing theoretical probability using playing cards (preview only - some features are disabled)

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