reviewing theoretical probability using playing cards (preview only - some features are disabled)
DESCRIPTION
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). - PowerPoint PPT PresentationTRANSCRIPT
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Reviewing THEORETICAL PROBABILITY
usingPLAYING CARDS
(PREVIEW ONLY - some features are disabled)
QUIZLESSONINTRO
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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!
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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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Because there are 4 kings in a deck of cards (1 of each suit), then 4 is the number of ways E can occur.
Probability of an event
The probability of an event, E, is:P(E) = number of ways E can occur
total number of outcomes
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Because there are 52 cards, then 52 is the total number of outcomes (of drawing a card).
Probability of an event
The probability of an event, E, is:P(E) = number of ways E can occur
total number of outcomes
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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.
Probability of an event
The probability of an event, E, is:P(E) = number of ways E can occur
total number of outcomes
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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.
Probability of an event NOT occurring
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.
Probability of an event NOT occurring
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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“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)
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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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“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)
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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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“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)
LAST
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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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“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)
LAST
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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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“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 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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“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)
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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
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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:
“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)
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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), 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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“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)
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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.
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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), writtenP(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), writtenP(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), writtenP(A and B) = P(A) X P(B|A)
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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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“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)
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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.
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CREDITS
HOME
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