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Probability

Probability. At the end of the period, you should be able to:. State the formula for the probability of any event A. Determine the probability of an event. Derive the formula for the probability of an event using the examples as guide.

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Probability

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  1. Probability

  2. At the end of the period, you should be able to: • State the formula for the probability of any event A. • Determine the probability of an event. • Derive the formula for the probability of an event using the examples as guide. • Solve at least 3 out of 5 word problems involving probability.

  3. What is Probability? Probability is something that each one of you is familiar with, and it can be shown by asking you this question: In rolling a regular six-sided die, what are the chances that you’d get 1 as its outcome? Answer: P(1) = 1 out of 6

  4. As you can see, all of you know the answer. So what is probability then? Probability is the exact value of chance that an event will happen. In order to easily understand probability, you have to be familiar with a couple of terms.

  5. Experiment - any process that generates outcomes and can be done repeatedly (e.g., rolling a die, flipping a coin, etc.) Outcome - the result in the experiment Sample Space - the set that contains all the possible outcomes of an experiment Event - any subset of the sample space - any collection of outcomes of an experiment

  6. Experiment: Rolling a die What does the number in the denominator represent? total # of possible outcomes What does the number in the numerator represent? # of outcomes that favors (satisfies) the event. P(1) = P(odd number) = P(even number) = P(4 or 5) =

  7. Experiment: Rolling a die If we let N = total number of possible outcomes n = number of outcomes that favors (satisfies) the event what will be the formula for the probability of any event E? P(E) =

  8. Our succeeding exercises/examples will deal with an ordinary deck of cards. To make sure that everyone knows the exact composition of the deck, here’s a summary: 52 cards, distributed equally (13)into 4 suits: diamonds, hearts, clubs (clovers), spades Each suit consists of A, 2, 3, 4, ..., 10, J, Q, K Letter cards - A, J, Q, K Face cards – J, Q, K Number cards – 2, 3, 4, 5, 6, 7,8 , 9, 10 Black cards - spades and clubs Red cards – hearts and diamonds

  9. Experiment: randomly selecting a card from a deck P(ace of diamond) = P(ace of spades) = P(diamond face card) = P(diamond) =

  10. Seatwork: ¼ sheet In the experiment of randomly selecting a card from a deck, solve/determine the probabilities of the ff: • P(black card) • P(face card) • P(number card) • P(letter card) • P(J or Q)

  11. Seatwork answers • P(black card) = • P(face card) = • P(number card) = • P(letter card) = • P(J or Q) =

  12. Day 2

  13. At the end of the period, you should be able to: • State the formula for the probability of any event A. • Solve for the total number of possible outcomes in (N)a given experiment. • Solve for the number of favorable outcomes (n) of a given event. • Solve for the probability of an event using the formula P(A) = n/N • Solve at least 3 out of 5 word problems involving probability.

  14. Recall: The formula used to determine the probability of any event A is given by: P(A) = n/N where n = number of favorable outcomes N = total number of possible outcomes in the experiment

  15. The Experiment Consider the experiment of drawing 5 cards from a deck of 52 cards. The total number of possible outcomes is: N = 52C5 = 2,598,960

  16. Example 1 What is the probability of having the cards to be all hearts? Solution: For the cards to be all hearts, we must choose only from the 13 heart cards. In how many ways can you choose 5 from the 13 heart cards? 13C 5 = 1,287 different ways

  17. Example 1 In randomly selecting 5 cards from a deck, what is the probability of having the cards to be all hearts? Solution: P(E) = n/N since n = 13C5 and N = 52C5 , P(all are hearts) = 13C5 / 52C5

  18. The cards in Example 1 is called a flush. Term to Remember: FLUSH - 5 cards of the same suit (e.g., flush of diamonds  all 5 cards are diamonds) (e.g., flush of hearts  all 5 cards are hearts) (e.g., flush of spades  all 5 cards are spades) (e.g., flush of clubs  all 5 cards are clubs)

  19. Example 2 What is the probability of getting a flush? Solution: N = 52C5 -four different possible flushes (1 for each suit) -in each suit, there are 13C5 ways of getting a flush thus, n = 13C5 x 4 (since there are four cases) P(flush) = 4 [13C5] / 52C5 The number of possible flushes (based on the different suits)

  20. Example 3 What is the probability of getting 3 aces & 2 kings? Solution: we need to choose: 3 aces  4C3 = 4 2 kings 4C2 = 6 Thus, n = 4C3 x 4C2 = 4 x 6 = 24 P(3 aces, 2 kings) = [4C3 x 4C2] / 52C5

  21. Term to Remember: FULLHOUSE - consist of three of a kind & a pair Examples: 3 aces and 2 queens 3 kings and 2 aces 3 jacks and 2 fours

  22. Example 4 What is the probability of getting a full house of aces? Solution: we need to choose: 3 aces  4C3 = 4 any pair  4C2, 12 possible pairs = 6 x 12 = 72 thus, n = 4C3 x (4C2 x 12) = 4 x 72 = 288 P(full house of aces) = [4C3 x 12(4C2)] / 52C5 The number of possible pairs (based on the different kinds)

  23. Example 5 What is the probability of getting 3 aces? Solution: we need to choose: 3aces  4C3 = 4 and 2 other cards (any 2 non-ace cards)  48C2 = 1,128 thus, n = 4C3 x 48C2 = 4 x 1128 = 4512 P(3 aces) = (4C3 x 48C2) / 52C5 The 2 other cards can be any card

  24. Terms to Remember: THREE OF A KIND – three cards of the same kind and 2 other different cards FOUR OF A KIND – four cards of the same kind and one other card ONE PAIR – one pair and three other cards that are all different

  25. Example 6 What is the probability of getting a three of a kind? Solution: need to choose: 3 cards of the same kind: 4C3 x 13 (there are 13 different cases)= 52 2 different cards, different from the 3 of a kind: 48C1 x 44C1 = 2,112 P(three of a kind) = [13(4C3) x 48C1 x 44C1] / 52C5 number of possible three of a kinds

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