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

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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  1. Reviewing THEORETICAL PROBABILITYusingPLAYING CARDS(PREVIEW ONLY - some features are disabled) INTRO LESSON QUIZ

  2. 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! HOME NEXT LAST

  3. 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 HOME NEXT LAST

  4. 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

  5. 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

  6. 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

  7. 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 HOME PREV NEXT LAST

  8. 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

  9. 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

  10. “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 HOME PREV NEXT LAST

  11. “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

  12. “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

  13. “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

  14. “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

  15. “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? HOME PREV NEXT LAST

  16. “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. HOME PREV NEXT LAST

  17. “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. HOME PREV NEXT LAST

  18. “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. HOME PREV NEXT LAST

  19. “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 HOME PREV NEXT LAST

  20. “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? HOME PREV NEXT LAST

  21. “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. HOME PREV NEXT LAST

  22. “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. HOME PREV NEXT LAST

  23. “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. HOME PREV NEXT LAST

  24. “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. HOME PREV NEXT LAST

  25. “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. HOME PREV NEXT LAST

  26. 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 HOME

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