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Binomial Probability


Binomial Probability

ARE YOU TELEPATHIC?

BigSuits

Can you read minds?
Can you read a computer's mind?
Does a computer have a mind?
Can you tell what the next card is?
Can you guess if it's a Heart, Club, Diamond, or Spade?
Concentrate, Click and See...


Which of the following is the better bet? If a coin is tossed four times, would you bet that it would turn up heads twice and tails twice or would you bet that it would not?

If a coin is tossed once, we will assume that the outcomes of turning up heads or tails are equally likely. Each outcome, then, has a probability of 1/2.

If the coin is tossed twice, there are four equally likely outcomes. They are illustrated in the tree diagram below. The probability that the coin will come up heads once and tails once is 50% because there are two ways out of the four of this happening and 2/4 = 1/2 = 50%.



Continuing the tree diagram to illustrate four tosses, we find that there are 16 equally likely outcomes.



Of these 16 possible outcomes, there are exactly 6 in which the coin turns up heads twice and tails twice. This means that the probability of this occurring is

6/16 = 3/8 = 37.5%

The probability that the coin will not turn up heads twice and tails twice is

10/16 = 5/8 = 62.5%

So betting that a coin will not turn up heads twice and tails twice when it is tossed four times is the better bet.

The chances of getting heads or tails when a coin is tossed is one example of binomial probability.

The probabilities in a situation in which there are two possible outcomes for each part are called binomial.

The two possible outcomes in binomial probability do not need to be equally likely. For example, suppose that you are playing a game in which the probability of winning is 4/5; that is, you can expect to win four out of every five games you play. It follows that your probability of losing is 1/5.



Now suppose that you play the game twice. The four possible outcomes, shown in this tree diagram, are not equally likely. The probability that you will win both games is 64%. The probability that you will lose both games is 4%. Because there are two outcomes in which you win exactly one game, each having a probability of 16%, the probability that you win exactly one game is

16% + 16% = 32%

These results are summarized in the table below.


    Number of games won     0   1   2
    Number of possibilities 1   2   1
    Percent probability     4%  32% 64%






EXERCISES — PART 1





  1. The tree diagram above shows the possibilities of boys and girls in a family with two children. According to birth records in the United States, the probability that a child is a boy is 0.512 and the probability that a child is a girl is 0.488. However, to keep calculations simple, we will assume that the chances of a child being a boy or girl are equally likely.
    1. Show why, if the probability of a child being a boy is 0.50, or 50%, the probability of two boys in a row is 0.25, or 25%.
    2. Show why the probability of a boy followed by a girl is also 25%. The diagram shows that there are two ways in which the family can consist of one boy and one girl: BG and GB, so the probability of a boy and a girl is 25% + 25% = 50%.
    3. Refer to the diagram to copy and complete the following table.
      
          Number of boys          0        1       2
          Number of possibilities 1
          Percent probability     25% 
      
      
      What is the percent probability that, in a family with two children,
    4. both children are of the same sex?
    5. at least one child is a girl?







  2. The tree diagram above shows the possibilities for three successive times that a baseball player comes up to bat. (H stands for hit and O for out.) We will assume that the player’s batting average is 0.400, so that the probability of a hit each time at bat is 0.4 and the probability of no hit is .6.
    1. Show why the probability of the player getting three hits in three times at bat, shown in the diagram as HHH, is 0.064, or 6.4%.
    2. Show why the probability of the player getting two hits followed by a miss, shown in the diagram as HHO, is 0.096, 9.6%. The diagram shows that there are three ways in which the player can exactly two hits: HHO, HOH, and OHH. Thus the probability of getting exactly two hits is 9.6% + 9.6% + 9.6% = 28.8%.
    3. What is the probability of the player getting a hit followed by two misses?
    4. How many ways does the diagram show in which the player gets exactly one hit?
    5. What is the probability of the player getting exactly one hit?
    6. What is the probability of the player getting no hits?
    7. Use the diagram and your answers above to copy and complete the following table.
      
      Number of hits              0   1   2   3
      Number of possibilities     1 
      Percent probability 
      
      
    8. Why should the four numbers in the second line of your table add up to 8?
    9. What should the four numbers in the third line of your table add up to? Use your table to answer the following questions.
    10. What is the most probable number of hits the player will get in three times at bat?
    11. What is the least probable number of hits? The probability that the player will get more than one hit is 28.8% + 6.4% = 35.2%
    12. What is the probability that the player will get fewer than two hits?







  3. The tree diagram above shows the possibilities of its snowing during four consecutive winter days at a ski resort. (S stands for snow and O for no snow.) Suppose that the probability of snow on any given day is 70%, 0.70, so that the probability of no snow is 30%, or 0.30.
    1. Show why the probability of its snowing all 4 days, shown in the diagram as SSSS, is 0.2401, or 24.01%.
    2. Show why the probability of snow on the first three days followed by no snow on the fourth day, shown in the diagram as SSSO, is 0.1029, or 10.29%.
    3. In how many ways does the diagram show it can snow on exactly three days?
    4. Show why the probability that it will snow on exactly three days is 41.16%.
    5. What is the probability that it will snow on the first two days, followed by no snow on the third and fourth days, shown in the figure as SSOO?
    6. In how many ways does the diagram show it can snow on exactly two days?
    7. What is the probability that it will snow on exactly two days?
    8. What is the probability that it will snow on the first day, followed by no snow on the other three days?
    9. In how many ways does the diagram show it can snow on exactly one day?
    10. What is the probability that it will snow on exactly one day?
    11. What is the probability that it will not snow on any of the four days?
    12. Use your answers above to copy and complete the following table. Round each probability to the nearest whole number.
      
          Number of days it snows 0   1   2   3   4
          Number of possibilities 1 
          Percent probability
      
      
    13. What should the five numbers in the second line of your table add up to?
    14. What should the five numbers in the third line of your table add up to?
    15. What is the most likely number of days that it will snow?
    16. What is the probability that it will snow on more than two days?





TELEPATHY — PART 2



BigSuits


INSTRUCTIONS

  1. Pick a partner.
  2. Select one of you as sender, and the other as receiver.
  3. The sender removes his or her glasses.
  4. The sender shuffles your deck of cards.
  5. The sender peeks randomly at one the cards without letting the receiver see it.
  6. The sender pictures the SUIT of the card in his or her mind for 10 seconds.
  7. The receiver attempts to read the mind of the sender.
  8. The receiver writes down which suit was received : CLUB DIAMOND HEART SPADE.
  9. The sender then writes down which card was sent, but does not reveal what is written.
  10. Then the sender returns the card to the deck and reshuffles all 52.
  11. Repeat this process four more times, which is five times in all.
  12. After all 5 cards are recorded, the sender shows the receiver the list of 5 cards.
  13. The receiver checks the correctly guessed suits on his or her list.
  14. The sender and receiver then switch roles.
  15. Let the sender become the receiver; and the receiver, the sender.
  16. Repeat the process described above for 5 more cards.
  17. Write your 5 guesses across the top of your paper.
  18. Circle those you guessed correctly.
  19. Under the correct guesses, write a W.
  20. Under the incorrect guesses, write an L.
  21. Write the probability for guessing correctly under the Winners.
  22. Write the probability for guessing incorrectly under the Losers.
  23. Answer all 10 questions below.
  24. Fill in the table and copy it neatly to your paper.
  25. Circle the probability in the table for correctly guessing your number of suits.
  26. Failure to follow these instructions will lose you points.


EXPECTED RESULTS

  1. What is the probability of guessing the suit of a card?
  2. What is the probability of not guessing the suit of a card?
  3. How many ways total can you guess the suits of 5 cards in this experiment?
  4. Approximately how many correct guesses should you expect if you are not telepathic?
    Hint: Examine the table.
  5. Fill in Column 1 with the answers to the following questions:
    Hint: Remember your coin flip assignment and Aces lab?
    1. With 5 cards, how many ways can you guess none of the suits?
    2. With 5 cards, how many ways can you guess exactly 1 suit correctly?
    3. With 5 cards, how many ways can you guess exactly 2 suits correctly?
    4. With 5 cards, how many ways can you guess exactly 3 suits correctly?
    5. With 5 cards, how many ways can you guess exactly 4 suits correctly?
    6. With 5 cards, how many ways can you guess exactly 5 suits correctly?


    FILL OUT THE COLUMNS IN THE TABLE BELOW

    
    x               = Number of correct guesses
    
    5Cx             = Count the ways x correct guesses can occur in 5 tries
                      Hint: Remember your coin flip assignment and Aces lab?
                      
    p^x = p(Suit)^x = Successive probability of guessing a suit correctly x times in 5 tries
    
    (1-p)^(5-x)     = Successive probability of guessing a suit incorrectly 5-x times in 5 tries
    p(~Suit)^(5-x)  
    
    P(x=#Suits)     = Probability of guessing a suit correctly x times in 5 tries
    
    



    OBSERVED RESULTS

  6. Were your guesses dependent or independent of each other? Why?
  7. What is the probability of guessing at most your number correctly?
  8. What is the probability of guessing at least your number correctly?
  9. Are you psychic?
    If the probability of guessing your number correctly is less than 10%,
    then the answer is YES;
    otherwise, answer NO!
  10. Write a formula that predicts the probability of correct guesses P(x) for any n number of cards.
    Use n, x, nCx, and p for symbols.
    P(x) = ?

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Last Modified 10/31/08 10:49 AM

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