Spring 2002

Selected solutions to review #1 questions

1. Solutions to some of the questions about poker hands.
      
• Exercise 2.5.5 The order in which the cards are received does not matter.
  Hence, the number of (five card) poker hands is
   
• Exercise 2.5.6 For each suit, there are ways to choose 5 cards from that suit. Since there are 4 suits, the total number of flushes is .
     
• Exercise 2.5.7 The lowest straight is 2-3-4-5-6 and the highest one is 10-J-Q-K-A. So, the highest card can be 6, 7, 8, 9, 10, J, Q, K, or A. This gives 9·4 = 36 choices for the highest card. Once that is chosen, there are 4 choices for each of the other cards -- each card has a definite "numerical" value, but there are 4 choices for its suit. So, there are 36·4⁴ = 9·4⁵ possible straights.
     
• Exercise 2.5.8 For a straight flush, there are again 36 choices for the highest card, but then only 1 choice for each of the remaining cards -- since all must be of the same suit as the highest card. Hence, there are 36·1⁴ = 36 straight flushes.
     
2. Solutions to some exercises from chapter 7 of the text.
      
• Exercise 7.1.10 For each type of hand, the probability is the number of ways to get that hand, divided by the total number of poker hands, namely, .
1. We use the result of exercise 2.5.6. Thus, the probability of being dealt a flush is  p = .
   Using the "invert and multiply" method with the compound fraction, we find that
   p = = = .00198.
      
   If we just compute the binomial coefficients numerically, it is:
   p = = .00198
      
2. Here, we use the result of exercise 2.5.7. The probability of being dealt a straight is  p = .
   By the same sort of calculation as before, we find that:
   p = = = = .00355
      
3. This time, we have  p =  , which is smaller than the previous answer by a factor of 4⁴ = 256.
   So,  p = .000014
      
• Exercise 7.3.9
  1. There are 5·4 = 20 ways to get a walk and a single. (The counting part of this is a permutation problem.) So there are 5 possibilities for choosing which time at bat the player gets a walk; for each of those, there are 4 possibilities for choosing the time at bat when s/he gets a single. For each of those individual possibilities, the probability is 0.1·0.2·0.6 = .012. So, p = .012·20 = .24.
        
  2. The probability of not getting put out in a given at-bat is 0.4. (You can calculate this as 1 - 0.6 or as 0.1 + 0.2 + 0.1.) So the probability of a perfect day is (0.4)⁵ = .01024, or about a 1% probability.
        
  3. We're assuming that the outcomes of the different times at bat are independent.
     This is roughly true but doesn't account for changes in mood (depending on how well previous times at bat have gone), on the pitcher and batter figuring out stuff about what kind of day the other is having, etc.
        
• Exercise 7.6.6 While this problem is "not hard", and can be done without using any of our major formulas, we still have to be careful if we want to get the right answer.
  • Step 1 Recall from the background info that there are 180 ways to arrange the letters of the word READER. So the answer is going to be a fraction whose denominator is 180.
     
  • Step 2 We have to count the number of these words in which the two E's are adjacent. Then there are 5 possible places where we can put the two E's. {Indeed, the first of the two adjacent E's can be the first, second, third, fourth or fifth letter of the word, but not the sixth letter.} For each of these 5 possibilities, we have to figure out how many possibilities there are for positioning the two R's [which don't have to be adjacent] and the remaining two letters.
     
    Thus, for each of the 5 possibilities, there are four spots not occupied by the E's. We can choose two of these four spots for the R's (not required to be adjacent). This can be done in  =  ways. For each of the possible ways to place the E's and the R's, we can place the remaining 2 letters (A and D) in 2! = 2 ways. So, altogether there are 5·6·2 = 60 words formed from the letters of the word READER in which the two E's are adjacent.
     
  • Step 3 We just divide: the answer is     p= 60/180 = 1/3.
  
   
• Exercise 7.6.7   
  1. We're supposed to find the probability that the two E's are adjacent, given that the first letter is an R. The answer is obtained by dividing the number of ways in which we can form a word whose first letter is R and the two E's are adjacent, divided by the total number of words in which the first letter is R.
     • Step 1: Among the 5 non-initial letters, there are  = 10 ways to place the two E's, and then there are 3! = 6 ways to arrange the remaining 3 letters (A, D, and the other R). So, there are 6·10 = 60 words in which the first letter is an R.
         
     • Step 2: When we require that the E's be adjacent and that the first letter be an R, then there are 4 ways to place the two adjacent Es' among the 5 non-initial letters, and then 3! = 6 ways to arrange the remaining 3 letters (A, D, and the other R). So, there are 4·6 = 24 words in which the first letter is an R and the two E's are adjacent.
         
     • Step 3: The answer is  p=24/60 = 2/5 = .4.
         
     Thus, if we know that the word starts with an R, this makes it more likely that the two E's are adjacent.
      
  2. We're supposed to find the probability that the two E's are adjacent, given that the first letter is an E. The general strategy is parallel to what we did in part i.
     • Step 1: This is just like Step 1 of part i. above, except that the roles of the E's and the R's are reversed. So, we find that there are 60 words in which the first letter is an E.
         
     • Step 2. Since the two E's have to be adjacent, there's just 1 spot to place the non-initial E, namely in the second spot. Now, the number of ways to place the two R's is  = 6, and the number of ways to arrange the remaining 2 letters is 2! = 2. So, there are just 6·2 = 12 words in which the first letter is an E and the two E's are adjacent.
       
     • Step 3: The answer is  p=12/60 = 1/5 = .2.
         
     Thus, if we know that the word starts with an E, this makes it less likely that the two E's are adjacent.
      
  3. The method for this and for part iv is quite similar
       
3. Solutions to the miscellaneous problems.  
    
   1. This is a conditional probability problem.
      1.  
         • Method 1: From our frequency chart for eye color, we know that there are 15 students whose eyes are not blue. By going back to Table 7.2 on page 149 of the text, we see that 10 of these students are female. So, the (conditional) probability is 8/15.
            
         • Method 2: We refer to Table 7.1 in section 7.6 of the text. The branch at the far right at the lower level shows that the probability is 8/15. (This method is not essentially different from method 1,because these probabilities were found by counting the various students in Table 7.2.
      2. There are 12 female students, of whom 8 are not blue eyed. So the conditional probability is 8/12 = 2/3. If you're doing this by studying the diagrams, you should look at figure 7.2 on page 153.
          
   2. In working parts a, b, c, and d, we note that there are = 22,100 ways to choose 3 cards from a deck of 52 cards. This is the same as the number of 3-element subsets of a 52-element set. (See chapter 2. If there is any remaining uncertainty about this, then review that material right away.)
        
      1. There are = 286 ways to choose 3 diamonds from the deck. So, the probability that all 3 or our cards are diamonds is  = .0129. To express this with products and quotients of binomial coefficients, it is .
      2. Since there are 26 red cards, there are = 2600 ways to choose 3 red cards from the deck. So, the probability that all 3 or our cards are red is  = .1176. To express this with products and quotients of binomial coefficients, it is .
      3. There are 4·3 = 12 face cards, so that the number of ways to choose 3 of them is  = 220. So, the probability that 3 face cards are chosen is  = .00995. To express this with products and quotients of binomial coefficients, it is
      4. Let E be "all 3 cards are diamonds", and let F be "all 3 cards are red". So, by part b, we have  p(F) = .1176. Since diamonds happen to be red,  EF  still is the event "all 3 cards are red. So,  p(EF) = .0129. The conditional probability therefore is  = .1097.
      5. Let E be "all 3 cards are face cards", and let F be "all 3 cards are diamonds". So, by part b, we have  p(F) = .0129. Then  EF  is the event "all 3 cards are diamonds and also face cards". This amounts to being king of diamonds, the queen of diamonds, and the jack of diamonds So,  p(EF) = .000045. The conditional probability therefore is  = .00349.
      6. Here, your formula will have the same numerator as in part f, but the denominator will be .00995. So, the answer will be somewhat larger.

         

      Back to the top of this page

      Back to the review questions

      Back to the class homepage