Finite Math—Final Solutions 1) a Child Has 30 Coins in Her Piggy Bank

Finite Math|Final Solutions 1) A child has 30 coins in her piggy bank. The change, consisting of dimes, nickels, and quarters only, adds up to $2.75. Also, the number of nickels exceeds the sum of the number of quarters and dimes by 10 coins. How many nickels, quarters, and dimes does she have? (10 points) Let x, y, and z be the number of nickels, dimes, and quarters respectively. We need to solve the following system of equations. x + y + z = 30 :05x + :10y + :25z = 2:75 x − y − z = 10 Add the first and last equation to get 2x = 40, which means x = 20: Now substitute this value to obtain a simpler system. y + z = 10 :10y + :25z = 1:75 Now multiply the top row by :10 and then subtract the two rows. This gives us :15z = :75, which means z = 5: Since y + z = 10, y = 5: So there are 20 nickels, 5 dimes, and 5 quarters. 2) Find the minimum of the objective function z = 4x1 + 8x2 + 3x3 subject to the given constraints (10 points). 3x1 + 4x2 + 2x3 ≥ 12; x1 + 2x2 + x3 ≥ 5; x1 ≥ 0; x2 ≥ 0; x3 ≥ 0 The Dual Problem is to maximize w = 12y1 + 5y2 subject to the constraints 3y1 + y2 ≤ 4; 4y1 + 2y2 ≤ 8; 2y1 + y2 ≤ 3; y1 ≥ 0; y2 ≥ 0: We now carry out the simplex method for the dual problem; the slack variables are called x1, x2, and x3 here. y1 y2 x1 x2 x3 w Ratios 4 3 1 1 0 0 0 4 3 4 2 0 1 0 0 8 2 3 2 1 0 0 1 0 3 2 −12 −5 0 0 0 1 0 Finite Math Question 2 Summer 2009 y1 y2 x1 x2 x3 w Ratios 1 1 4 1 3 3 0 0 0 3 4 2 4 8 0 3 − 3 1 0 0 3 4 1 2 1 0 3 − 3 0 1 0 3 1 0 −1 4 0 0 1 16 y1 y2 x1 x2 x3 w 1 0 1 0 −1 0 1 0 0 0 1 −2 0 2 0 1 −2 0 3 0 1 0 0 2 0 3 1 17 So the minimum is 17 and it occurs at x1 = 2, x2 = 0, and x3 = 3: Since there are only two variables in the dual maximization problem, we can also draw the feasible region and test the corner points. The maximum of w is 17 and occurs at (1; 1): This means that the minimum of z is 17. Note that this solution does not provide the values of x1, x2, and x3 where the minimum occurs; as luck would have it, you are not being asked to provide these values. Finite Math Question 3 Summer 2009 3) A fair coin is tossed five time. Calculate the probability of the following events (10 points). a) The third toss results in Heads. 2 × 2 × 1 × 2 × 2 2 × 2 × 2 × 2 × 2 Another way to see this is to observe that each toss in independent of all the other tosses, 1 and so, the probability is 2 : b) At least 3 Tails appear. C(5; 3) + C(5; 4) + C(5; 5) 1 = 25 2 Another solution would be to consider the two mutually exclusive complementary events \At least 3 Heads" and \At least 3 Tails"|if there are less than 3 Heads, then there must be at least 3 Tails and vice versa. So P (At least 3 Heads) + P (At least 3 Tails) = 1: Also, these events are symmetric and have equal probabilities, that is, P (At least 3 Heads) = P (At least 3 Tails): 1 These two equations imply that P (At least 3 Heads) = P (At least 3 Tails) = 2 : This a \general fact" and there is nothing special about the numbers 5 and 3 above (except that 5 is odd). If the coin is tossed 3 times, then the probability of getting at least 2 tails is a half. If the coin is tossed 7 times, then the probability of getting at least 4 tails is a half. If the coin is tossed 9 times, the probability of getting at least 5 tails is a half... 4) A small manufacturing company makes two different kinds of spark plugs, wide-gap and narrow-gap plugs. Each wide-gap spark plug requires 1 hour of labor and costs $3 to produce, and each narrow-gap spark plug requires 4 hours of labor and costs $2 to produce. The number of labor hours available per day is not to exceed 40 hours and the total daily funds available for production are $60. If the wide-gap spark plugs sell for $4 and the narrow-gap spark plugs sell for $3, how many of each should be produced daily in order to maximize profit? (10 points) Let x and y be number of wide-gap and narrow-gap spark plugs respectively. We have to maximize z = (4x + 3y) − (3x + 2y) = x + y subject to x + 4y ≤ 40; 3x + 2y ≤ 60; x ≥ 0; y ≥ 0: Finite Math Question 4 Summer 2009 After drawing the feasible region, we test all the corner points and discover that the maximum is realized at (16; 6): So the company should manufacture 16 wide-gap and 6 narrow- gap spark plugs every day. (0; 0) :z = (0) + (0) = 0 (0; 10) :z = (0) + (10) = 10 (20; 0) :z = (20) + (0) = 20 (16; 6) :z = (16) + (6) = 22 Alternatively, this problem can be solved by employing the simplex method. x y s1 s2 z Ratios 1 4 1 0 0 40 40 3 2 0 1 0 60 20 −1 −1 0 0 1 0 x y s1 s2 z Ratios 10 1 0 3 1 − 3 0 20 12 2 1 1 3 0 3 0 20 30 1 1 0 − 3 0 3 1 20 Finite Math Question 5 and 6 Summer 2009 x y s1 s2 z 3 1 0 1 10 − 10 0 6 1 2 1 0 − 5 5 0 16 1 3 0 0 10 10 1 22 So the maximum is 22 and it is realized at x = 16 and y = 6: 5) Two cards are randomly drawn in succession without replacement from a standard deck of 52 cards (10 points). a) What is the probability that the first card is a face card (Jack, Queen, or King) given that the second card is not a face card? 12 × 40 12 52 × 51 = 51 × 40 51 52 × 51 b) Find the probability that at least one face card is selected given that both cards belong to the different suites. 2 × 12 × 30 + 12 × 9 52 × 51 52 × 39 52 × 51 6) A bag contains 2 pearls and nothing else. A fair die is rolled. If the outcome of the roll is 1 or 2, two white beads are added to the bag, if the outcome of the roll is a 3 or 4, four white beads are added to the bag, and if the roll is a 5 or 6, six white beads are added to the bag. Then an object is drawn at random from the bag. Given that a pearl was chosen, find the probability that 2 white beads were added to the bag (10 points). This is a standard Bayes' Rule problem. 1 2 × 3 4 1 2 1 2 1 2 × + × + × 3 4 3 6 3 8.

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