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IT 223 -- Feb 23, 2026

Review Exercises

1. An American roulette wheel contains 38 pockets: 18 red pockets, 18 black pockets, and two green pockets labeled 0 and 00. European roulette wheels have only one green pocket labeled 0.
   If you bet on red or black, here are the payoff tables for an American roulette wheel, for a bet of $100:
   

   Bet on Red
   Outcome	Payoff	Probability
   Red	+100	18/38 = 0.474
   Black	-100	18/38 = 0.474
   Green	-100	2/38 = 0.052

   
   

   Bet on Black
   Outcome	Payoff	Probability
   Red	-100	18/38 = 0.474
   Black	+100	18/38 = 0.474
   Green	-100	2/38 = 0.052

   
   What is the expected payoff of playing on red? playing on black?
   Answer: Playing on red is
        E(x) = 100(0.474) + (-100)(0.474) + (-100)(0.052) = -5.26.
   You expect to lose about $5.26 each time you play on red with a $100 bet; same for black.
2. The following boxes contain 6 tickets each. In which urns are letter independent from color? In other words, in which urns are the probabilities of the letters the same whether or not you specify the color?
   
   
   Answer: In a and b, letter is independent of number. To show that letter and number are independent, you need to show that for each letter value L, P(L)= P(L|C) for a specific color C. For example, in Problem 2a, if you are not restricted to a specific color,
   P(A) = P(B) = P(C) = 2/6 = 1/3.
   If you know the color is red, P(A|red) = P(B|red) = P(C|red) = 1/3. If you know the color is green, the probabilities are also 1/3.
   In Problem 8c, if you don't know the color, P(F) = 2/6 = 1/3. If you know that the color is green, P(F|green) = 2/3 = 2/3, so color and letter are not independent.
   To show that two events are not independent, it is enough to find one case where the probabilities are different; to show that two events are independent, you must show that the probabilities are the same for all choices of color and letter.
3. If you know the expected value of a random variable x, what is the expected value of the sum of n outcomes of this random variable x?
   Answer: E(S) = nE(x)
4. If you know the variance of a random variable x, what is the expected value of the sum of n independent outcomes of this random variable x?
   Answer: Var(S) = nVar(x)
5. What is the Law of Averages? What is the officlal name of the Law of Averages?
   Answer: The Law of Averages says that if the number of observations n is large, the average of n independent outcomes of a random variable x is close to the true expected value E(x) of the random variable.

Averages of Random Variables

• The expected value of the average of n outcomes of a random variable:
  E(x) = E[(x₁ + ... + x_n) / n] = E(x₁ / n + ... + x_n / n)
      = E(x₁ / n) + ... + E(x_n / n) = E(x₁) / n + ... + E(x_n) / n
      = nE(x) / n = E(x)
  

  E(x) = E(x)

• The variance of the sum of n independent outcomes of a random variable:
  Var(x) = Var[(x₁ + ... + x_n) / n] = Var[x₁ / n + ... + x_n / n]
      = Var(x₁ / n) + ... + Var(x_n / n) = Var(x₁) / n² + ... + Var(x_n) / n²
      = = nVar(x) / n² = nVar(x)
  

  Var(x) = nVar(x)

• The standard deviation of the sum of n independent outcomes of a random variable:
  σ_x = √[Var(x)] = √[Var(x) / n] = σ_x / √n
  

  σx = σx / √n

• Recall that E(x) = E(x). How does σ_x compare to σ_x?

The Central Limit Theorem

• Even if a random variable x is not normally distributed, both the sum S and average x of n independent outcomes of a random variable are approximately normally distributed if n is large. The larger n is, the closer S is to a normal distribution. If S is a sum of independent Bernoulli random variables with p = probability of obtaining a 1, then if np > 10 and n(1-p) > 10, approximating S by a normal distribution is generally valid.
• Abraham de Moivre first proved the Central Limit Theorem (CLT) in 1733.

Factorials and Counting Combinations

• The number of ways of drawing k objects from an urn that contains n distinct objects is
  
• The binomial coefficient defined in the previous bullet point is also written as nCk and read "n choose k."
• The binomial coefficients can also be obtained from Pascal's Triangle:
  

                      1
                   1     1
                1     2     1
              1    3     3     1
           1    4     6     4     1
        1     5    10    10    5     1
      1    6    15    20    15    6     1
    1   7    21    35    35    21    7     1
  1   8   28    56    70    56    28    8     1
  

• Each entry in Pascal's triangle can be obtained recursively by adding the two entries above it.
• nCk can be obtained from in Pascal's Triangle as entry k in row n.
• Practice Problems: How many ways are there of choosing
  
  1. 2 items from 3?
     Answer: to use Pascal's triangle start at the 0 at the top and count 0, 1, 2, 3 to row 3.  Then count 0, 1, 2 from the left.  The entry at that location is 3.
     Using the formula: 3C2 = 3! / 2!(3 - 2)! = 3*2*1 / ((2*1)*1) = 3.
     Listing directly the number of ways of choosing 2 items from 3: (12), (13), (23): 3 ways.
     Using R:
     

     > choose(3, 2)
     [1] 3
     

  2. 3 items from 5?
     Answer: Using Pascal's triangle, count 0 to 5 from the top, and count 0 to 3 from the left. The entry is 10.
     Using the formula: 5C3 = 5! / (3!(5 - 3)!) = (5*4*3*2*1) / ((3*2*1)(2*1) = 5*2 = 10.
     Listing directly, the number of ways of choosing 3 items from 5 is
     (123) (124) (125) (134) (135) (145) (234) (235) (245) (345):  10 ways.
     Using R:
     

     > choose(5, 3)
     [1] 10
     

  3. 5 items from 8?
     Answer: 8C5 = 8! / (5!(8 - 5)!) = (8*7*6*5*4*3*2*1) / ((5*4*3*2*1)(3*2*1)) = 8 * 7 = 56,
     Using R:
     

     > choose(8, 5)
     [1] 56
     

  4. 13 items from 20?
     Answer: 20C13 = 20! / (13!(20 - 13)!) = (20*19*18*17*16*15*14) / (7*6*5*4*3*2*1) = 19*17*16*15 = 77520,
     Using R:
     

     > choose(20, 13)
     [1] 77520
     

The Binomial Theorem

• Suppose you have a Bernoulli distribution with p = probability of success. What is the probability of k successes out of n trials? Answer:
  
  
  
• This formula can also be written as
  P(n successes in k trials) = nCk p^k (1 - p)^n-k or
       nCk * p^k * (1-p)^(n-k)
• Practice Problems:
  
  1. Flip 5 coins. What is the probability of obtaining exactly 3 heads?
     Answer: the probability is
     nCk p^k (1 - p)^n-k = 5C3 0.5⁵ (1 - 0.5)^{10 - 5} = 10! / (5!(10-5)!) * 0.125 * 0.25 = 0.3125.
     Using R:
     

      > dbinom(3, 5, 0.5)
     [1] 0.3125
     

  2. Roll a die 10 times. What is the probability of rolling exactly 5 ones?
     Answer: nCk p^k (1 - p)^n-k = 10C5 (1/6)⁵ (1 - 1/6)^10-5 =  
     10! / 5!(10-5)! *  0.0001286008 * 0.4018776 = (10*9*8*7*6)/(5*4*3*2*1) * 5.168178e-05 =
     (3*2*7*6) * 5.168178e-05 = 252 * 5.168178e-05 = 0.01302381.
     Using R:
     

     > dbinom(5, 10, 1/6)
     [1] 0.01302381
     

  3. Shoot 50 free throws, where the probability of success is 80%. What is the probability of hitting 45 or more free throws?
     Answer: perform the "hand" calculations with R:
     

     > choose(50, 45) * 0.7^45 * (1-0.7)^(50-45) +
     + choose(50, 46) * 0.7^46 * (1-0.7)^(50-46) +
     + choose(50, 47) * 0.7^47 * (1-0.7)^(50-47) +
     + choose(50, 48) * 0.7^48 * (1-0.7)^(50-48) +
     + choose(50, 49) * 0.7^49 * (1-0.7)^(50-49) +
     + choose(50, 50) * 0.7^50 * (1-0.7)^(50-50)
     [1] 0.0007228617
     > # Next compute these probabilities with dbinom:
     > dbinom(45, 50, 0.7) + dbinom(46, 50, 0.7) + dbinom(47, 50, 0.7) + 
     + dbinom(48, 50, 0.7) + dbinom(49, 50, 0.7) + dbinom(50, 50, 0.7)
     [1] 0.0007228617
     > # Finally pbinom(44, 50, 0.7) computes the probability of 
     > # making 44 or fewer free throws.  1 - pbinom(44, 50, 0.7) is
     > # the probability of making 45 or more free throws:
     > 1 - pbinom(44, 50, 0.7)
     [1] 0.0007228617

  4. The probability that the Cubs can beat Team A in a single game is 60%. What is the probability that they win 7 or more games in a 10 game series?
     Answer: Perform the calculation first using dbinom:
     

     > dbinom(7, 10, 0.6) + dbinom(8, 10, 0.6) + 
     + dbinom(9, 10, 0.6) + dbinom(10, 10, 0.6)
     [1] 0.3822806
     

     Next, perform it with pbinom:
     

     > 1 - pbinom(6, 10, 0.6)
     [1] 0.3822806
     

Finding a Confidence Interval for a Proportion

• Example 1:   Flip a coin 1,000 times and obtain 514 heads. Find a 95% confidence interval for the true value p of obtaining a head. This will give us an idea if the coin is fair.
  Ans: Flipping a coin once is a Bernoulli random variable with unknown probability of obtaining a head p. We also know that the sum S = 514, and the estimated probability of success
        p^{^} = S / n = 514 / 1,0000 = 0.514.
  We know that
        E(S) = nE(x) = np.
  Furthermore, we know that
        σ_S = √[np(1-p)].
  Since p is unknown we estimate it with p^{^} and obtain
      σ_S  = √[np^{^}(1-p^{^})] = √[1000(0.514)(1 - 0.514)] = 15.81.
  Finally, we compute the z-score
        z = (S - E(S)) / σ_S = (514 - np) / 15.81.
  To obtain z, we have scaled S by subtracting its expectation and dividing by its standard error.  This means that E(z) = 0 and σ_z = 1. By the Central Limit Theorem, z is a standard normal random variable if n > 30, so -2 ≤ z ≤ 2 95% of the time. (More precisely, -1.96 ≤ z ≤ 1.96 95% of the time. We can then solve this inequality for for p:
       -2 ≤ z ≤ 2
       -2 ≤ (S - np) / σ_S ≤ 2
       -2 ≤ (514 - 1,000p) / 15.81 ≤ 2
       -31.62 ≤ 514 - 1,000p ≤ 31.62
       -31.62 - 514 ≤ - 1,000p ≤ 31.62 - 514
       -545.62 ≤ - 1,000p ≤ -475.62
       0.54562 ≥ p ≥ 0.47562
  This means that the 95% confidence interval is
       [0.47562, 0.54562] or [48%, 55%].

Project 4

• Look at the Project 4 Description.
• Simulating repeated trials from a Bernoulli random variable using R:
       Method 1: Generate n values using rbinom(n, 1, p)
       Method 2: Generate one value using rbinom(1, n, p). The one value is the number of ones obtained, which is the sum of the outcomes.