6.2: Problems on Random Variables and Probabilities
Exercise \(\PageIndex{1}\)
The following simple random variable is in canonical form:
\(X = -3.75 I_A - 1.13 I_B + 0 I_C + 2.6 I_D\).
Express the events \(\{X \in (-4, 2]\}\), \(\{X \in (0, 3]\}\), \(\{X \in (-\infty, 1]\}\), and {\(X \ge 0\)} in terms of \(A\), \(B\), \(C\), and \(D\).
- Answer
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- \(A \bigvee B \bigvee C\)
- \(D\)
- \(A \bigvee B \bigvee C\)
- \(C\)
- \(C \bigvee D\)
Exercise \(\PageIndex{2}\)
Random variable \(X\), in canonical form, is given by \(X = -2I_{A} - I_B + I_C + 2I_D + 5I_E\).
Express the events \(\{X \in [2, 3)\}\), \(\{X \le 0\}\), \(\{X < 0\}\), \(\{|X - 2| \le 3\}\), and \(\{X^2 \ge 4\}\), in terms of \(A, B, C, D, and E\).
- Answer
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- \(D\)
- \(A \bigvee B\)
- \(A \bigvee B\)
- \(B \bigvee C \bigvee D \bigvee E\)
- \(A \bigvee D \bigvee E\)
Exercise \(\PageIndex{3}\)
The class \(\{C_j: 1 \le j \le 10\}\) is a partition. Random variable \(X\) has values {1, 3, 2, 3, 4, 2, 1, 3, 5, 2} on \(C_1\) through \(C_{10}\), respectively. Express X\) in canonical form.
- Answer
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T = [1 3 2 3 4 2 1 3 5 2]; [X,I] = sort(T) X = 1 1 2 2 2 3 3 3 4 5 I = 1 7 3 6 10 2 4 8 5 9
\(X = I_A + 2I_B + 3I_C + 4I_D + 5I_E\)
\(A = C_1 \bigvee C_7\), \(B = C_3 \bigvee C_6 \bigvee C_{10}\), \(C = C_2 \bigvee C_4 \bigvee C_8\), \(D = C_5\), \(E = C_9\)
Exercise \(\PageIndex{4}\)
The class \(\{C_j: 1 \le j \le 10\}\) in Exercise has respective probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine the distribution for \(X\)
- Answer
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T = [1 3 2 3 4 2 1 3 5 2]; pc = 0.01*[8 13 6 9 14 11 12 7 11 9]; [X,PX] = csort(T,pc); disp([X;PX]') 1.0000 0.2000 2.0000 0.2600 3.0000 0.2900 4.0000 0.1400 5.0000 0.1100
Exercise \(\PageIndex{5}\)
A wheel is spun yielding on an equally likely basis the integers 1 through 10. Let C i be the event the wheel stops at \(i\), \(1 \le i \le 10\). Each \(P(C_i) = 0.1\). If the numbers 1, 4, or 7 turn up, the player loses ten dollars; if the numbers 2, 5, or 8 turn up, the player gains nothing; if the numbers 3, 6, or 9 turn up, the player gains ten dollars; if the number 10 turns up, the player loses one dollar. The random variable expressing the results may be expressed in primitive form as
\(X = -10I_{C_1} + 0I_{C_2} + 10I_{C_3} - 10I_{C_4} + 0I_{C_5} + 10I_{C_6} - 10I_{C_7} + 0I_{C_8} + 10I_{C_9} - I_{C_{10}}\)
- Determine the distribution for \(X\), (a) by hand, (b) using MATLAB.
- Determine \(P(X < 0)\), \(P(X > 0)\).
- Answer
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p = 0.1*ones(1,10); c = [-10 0 10 -10 0 10 -10 0 10 -1]; [X,PX] = csort(c,p); disp([X;PX]') -10.0000 0.3000 -1.0000 0.1000 0 0.3000 10.0000 0.3000 Pneg = (X<0)*PX' Pneg = 0.4000 Ppos = (X>0)*PX' Ppos = 0.300
Exercise \(\PageIndex{6}\)
A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written
\(X = 3.5 I_{C_1} + 5.0 I_{C_2} + 3.5 I_{C_3} + 7.5 I_{C_4} + 5.0 I_{C_5} + 5.0I_{C_6} + 3.5 I_{C_7} + 7.5I_{C_8}\)
Determine the distribution for \(X\) (a) by hand, (b) using MATLAB.
- Answer
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p = 0.01*[10 15 15 20 10 5 10 15]; c = [3.5 5 3.5 7.5 5 5 3.5 7.5]; [X,PX] = csort(c,p); disp([X;PX]') 3.5000 0.3500 5.0000 0.3000 7.5000 0.3500
Exercise \(\PageIndex{7}\)
Suppose \(X\), \(Y\) in canonical form are
\(X = 2 I_{A_1} + 3 I_{A_2} + 5 I_{A_3}\) \(Y = I_{B_1} + 2 I_{B_2} + 3I_{B_3}\)
The \(P(A_i)\) are 0.3, 0.6, 0.1, respectively, and the \(P(B_j)\) are 0.2 0.6 0.2. Each pair {\(A_i, B_j\)} is independent. Consider the random variable \(Z = X + Y\). Then \(Z = 2 + 1\) on \(A_1 B_1\), \(Z = 3 + 3\) on \(A_2 B_3\), etc. Determine the value of \(Z\) on each \(A_i B_j\) and determine the corresponding \(P(A_i B_j)\). From this, determine the distribution for \(Z\).
- Answer
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A = [2 3 5]; B = [1 2 3]; a = rowcopy(A,3); b = colcopy(B,3); Z =a + b % Possible values of sum Z = X + Y Z = 3 4 6 4 5 7 5 6 8 PA = [0.3 0.6 0.1]; PB = [0.2 0.6 0.2]; pa= rowcopy(PA,3); pb = colcopy(PB,3); P = pa.*pb % Probabilities for various values P = 0.0600 0.1200 0.0200 0.1800 0.3600 0.0600 0.0600 0.1200 0.0200 [Z,PZ] = csort(Z,P); disp([Z;PZ]') % Distribution for Z = X + Y 3.0000 0.0600 4.0000 0.3000 5.0000 0.4200 6.0000 0.1400 7.0000 0.0600 8.0000 0.0200
Exercise \(\PageIndex{8}\)
For the random variables in Exercise, let \(W = XY\). Determine the value of \(W\) on each \(A_i B_j\) and determine the distribution of \(W\).
- Answer
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XY = a.*b XY = 2 3 5 % XY values 4 6 10 6 9 15 W PW % Distribution for W = XY 2.0000 0.0600 3.0000 0.1200 4.0000 0.1800 5.0000 0.0200 6.0000 0.4200 9.0000 0.1200 10.0000 0.0600 15.0000 0.0200
Exercise \(\PageIndex{9}\)
A pair of dice is rolled.
- Let \(X\) be the minimum of the two numbers which turn up. Determine the distribution for \(X\)
- Let \(Y\) be the maximum of the two numbers. Determine the distribution for \(Y\).
- Let \(Z\) be the sum of the two numbers. Determine the distribution for \(Z\).
- Let \(W\) be the absolute value of the difference. Determine its distribution.
- Answer
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t = 1:6; c = ones(6,6); [x,y] = meshgrid(t,t) x = 1 2 3 4 5 6 % x-values in each position 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 y = 1 1 1 1 1 1 % y-values in each position 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 m = min(x,y); % min in each position M = max(x,y); % max in each position s = x + y; % sum x+y in each position d = abs(x - y); % |x - y| in each position [X,fX] = csort(m,c) % sorts values and counts occurrences X = 1 2 3 4 5 6 fX = 11 9 7 5 3 1 % PX = fX/36 [Y,fY] = csort(M,c) Y = 1 2 3 4 5 6 fY = 1 3 5 7 9 11 % PY = fY/36 [Z,fZ] = csort(s,c) Z = 2 3 4 5 6 7 8 9 10 11 12 fZ = 1 2 3 4 5 6 5 4 3 2 1 %PZ = fZ/36 [W,fW] = csort(d,c) W = 0 1 2 3 4 5 fW = 6 10 8 6 4 2 % PW = fW/36
Exercise \(\PageIndex{10}\)
Minterm probabilities \(p(0)\) through \(p(15)\) for the class \(\{A, B , C, D\}\) are, in order,
0.072 0.048 0.018 0.012 0.168 0.112 0.042 0.028 0.062 0.048 0.028 0.010 0.170 0.110 0.040 0.
Determine the distribution for random variable
\(X = -5.3I_A - 2.5 I_B + 2.3 I_C + 4.2 I_D - 3.7\)
- Answer
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% file npr06_10.m % Data for Exercise 6.2.10. pm = [ 0.072 0.048 0.018 0.012 0.168 0.112 0.042 0.028 ... 0.062 0.048 0.028 0.010 0.170 0.110 0.040 0.032]; c = [-5.3 -2.5 2.3 4.2 -3.7]; disp('Minterm probabilities are in pm, coefficients in c') npr06_10 Minterm probabilities are in pm, coefficients in c canonic Enter row vector of coefficients c Enter row vector of minterm probabilities pm Use row matrices X and PX for calculations Call for XDBN to view the distribution XDBN XDBN = -11.5000 0.1700 -9.2000 0.0400 -9.0000 0.0620 -7.3000 0.1100 -6.7000 0.0280 -6.2000 0.1680 -5.0000 0.0320 -4.8000 0.0480 -3.9000 0.0420 -3.7000 0.0720 -2.5000 0.0100 -2.0000 0.1120 -1.4000 0.0180 0.3000 0.0280 0.5000 0.0480 2.8000 0.0120
Exercise \(\PageIndex{11}\)
On a Tuesday evening, the Houston Rockets, the Orlando Magic, and the Chicago Bulls all have games (but not with one another). Let A be the event the Rockets win, \(B\) be the event the Magic win, and \(C\) be the event the Bulls win. Suppose the class {\(A, B, C\)} is independent, with respective probabilities 0.75, 0.70 0.8. Ellen's boyfriend is a rabid Rockets fan, who does not like the Magic. He wants to bet on the games. She decides to take him up on his bets as follows:
- $10 to 5 on the Rockets --- i.e. She loses five if the Rockets win and gains ten if they lose
- $10 to 5 against the Magic
- even $5 to 5 on the Bulls.
Ellen's winning may be expressed as the random variable
\(X = -5 I_A + 10 I_{A^c} + 10 I_B - 5 I_{B^c} - 5 I_C + 5I_{C^c} = -15 I_A + 15 I_B - 10 I_C + 10\)
Determine the distribution for \(X\). What are the probabilities Ellen loses money, breaks even, or comes out ahead?
- Answer
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P = 0.01*[75 70 80]; c = [-15 15 -10 10]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) -15.0000 0.1800 -5.0000 0.0450 0 0.4800 10.0000 0.1200 15.0000 0.1400 25.0000 0.0350 PXneg = (X<0)*PX' PXneg = 0.2250 PX0 = (X==0)*PX' PX0 = 0.4800 PXpos = (X>0)*PX' PXpos = 0.2950
Exercise \(\PageIndex{12}\)
The class {\(A, B, C, D\)} has minterm probabilities
\(pm = 0.001 *\) [5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]
- Determine whether or not the class is independent.
- The random variable \(X = I_A + I_B + I_C + I_D\)counts the number of the events which occur on a trial. Find the distribution for X and determine the probability that two or more occur on a trial. Find the probability that one or three of these occur on a trial.
- Answer
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npr06_12 Minterm probabilities in pm, coefficients in c a = imintest(pm) The class is NOT independent Minterms for which the product rule fails a = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 canonic Enter row vector of coefficients c Enter row vector of minterm probabilities pm Use row matrices X and PX for calculations Call for XDBN to view the distribution XDBN = 0 0.0050 1.0000 0.0430 2.0000 0.2120 3.0000 0.4380 4.0000 0.3020 P2 = (X>=2)*PX' P2 = 0.9520 P13 = ((X==1)|(X==3))*PX' P13 = 0.4810
Exercise \(\PageIndex{13}\)
James is expecting three checks in the mail, for $20, $26, and $33 dollars. Their arrivals are the events \(A, B, C\). Assume the class is independent, with respective probabilities 0.90, 0.75, 0.80. Then
\(X = 20 I_A + 26 I_B + 33 I_C\)
represents the total amount received. Determine the distribution for \(X\). What is the probability he receives at least $50? Less than $30?
- Answer
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c = [20 26 33 0]; P = 0.01*[90 75 80]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) 0 0.0050 20.0000 0.0450 26.0000 0.0150 33.0000 0.0200 46.0000 0.1350 53.0000 0.1800 59.0000 0.0600 79.0000 0.5400 P50 = (X>=50)*PX' P50 = 0.7800 P30 = (X <30)*PX' P30 = 0.0650
Exercise \(\PageIndex{14}\)
A gambler places three bets. He puts down two dollars for each bet. He picks up three dollars (his original bet plus one dollar) if he wins the first bet, four dollars if he wins the second bet, and six dollars if he wins the third. His net winning can be represented by the random variable
\(X = 3I_A + 4I_B + 6I_C - 6\), with \(P(A) = 0.5\), \(P(B) = 0.4\), \(P(C) = 0.3\)
Assume the results of the games are independent. Determine the distribution for \(X\).
- Answer
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c = [3 4 6 -6]; P = 0.1*[5 4 3]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution dsp(XDBN) -6.0000 0.2100 -3.0000 0.2100 -2.0000 0.1400 0 0.0900 1.0000 0.1400 3.0000 0.0900 4.0000 0.0600 7.0000 0.0600
Exercise \(\PageIndex{15}\)
Henry goes to a hardware store. He considers a power drill at $35, a socket wrench set at $56, a set of screwdrivers at $18, a vise at $24, and hammer at $8. He decides independently on the purchases of the individual items, with respective probabilities 0.5, 0.6, 0.7, 0.4, 0.9. Let \(X\) be the amount of his total purchases. Determine the distribution for \(X\).
- Answer
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c = [35 56 18 24 8 0]; P = 0.1*[5 6 7 4 9]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) 0 0.0036 8.0000 0.0324 18.0000 0.0084 24.0000 0.0024 26.0000 0.0756 32.0000 0.0216 35.0000 0.0036 42.0000 0.0056 43.0000 0.0324 50.0000 0.0504 53.0000 0.0084 56.0000 0.0054 59.0000 0.0024 61.0000 0.0756 64.0000 0.0486 67.0000 0.0216 74.0000 0.0126 77.0000 0.0056 80.0000 0.0036 82.0000 0.1134 85.0000 0.0504 88.0000 0.0324 91.0000 0.0054 98.0000 0.0084 99.0000 0.0486 106.0000 0.0756 109.0000 0.0126 115.0000 0.0036 117.0000 0.1134 123.0000 0.0324 133.0000 0.0084 141.0000 0.0756
Exercise \(\PageIndex{16}\)
A sequence of trials (not necessarily independent) is performed. Let \(E_i\) be the event of success on the \(i\)th component trial. We associate with each trial a "payoff function" \(X_i = aI_{E_i} + b I_{E_i^c}\). Thus, an amount \(a\) is earned if there is a success on the trial and an amount \(b\) (usually negative) if there is a failure. Let \(S_n\) be the number of successes in the \(n\) trials and \(W\) be the net payoff. Show that \(W = (a - b) S_n + bn\).
- Answer
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\(X_i = aI_{E_i} + b(1 - I_{E_i}) = (a - b) I_{E_i} + b\)
\(W = \sum_{i = 1}^{n} X_i = (a - b) \sum_{i = 1}^{n} I_{E_i} + bn = (a - b) S_n + bn\)
Exercise \(\PageIndex{17}\)
A marker is placed at a reference position on a line (taken to be the origin); a coin is tossed repeatedly. If a head turns up, the marker is moved one unit to the right; if a tail turns up, the marker is moved one unit to the left.
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Show that the position at the end of ten tosses is given by the random variable
\(X = \sum_{i = 1}^{10} I_{E_i} - \sum_{i = 1}^{10} I_{E_i^c} = 2 \sum_{i = 1}^{10} I_{E_i} - 10 = 2S_{10} - 10\)
where \(E_i\) is the event of a head on the \(i\)th toss and \(S_{10}\) is the number of heads in ten trials.
- After ten tosses, what are the possible positions and the probabilities of being in each?
- Answer
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\(X_i = I_{E_i} - I_{E_i^c} = I_{E_i} - (1 - I_{E_i}) = 2I_{E_i} - 1\)
\(X = \sum_{i = 1}^{10} X_i = 2\sum_{i = 1}^{n} I_{E_i} - 10\)
S = 0:10; PS = ibinom(10,0.5,0:10); X = 2*S - 10; disp([X;PS]') -10.0000 0.0010 -8.0000 0.0098 -6.0000 0.0439 -4.0000 0.1172 -2.0000 0.2051 0 0.2461 2.0000 0.2051 4.0000 0.1172 6.0000 0.0439 8.0000 0.0098 10.0000 0.0010
Exercise \(\PageIndex{18}\)
Margaret considers five purchases in the amounts 5, 17, 21, 8, 15 dollars with respective probabilities 0.37, 0.22, 0.38, 0.81, 0.63. Anne contemplates six purchases in the amounts 8, 15, 12, 18, 15, 12 dollars, with respective probabilities 0.77, 0.52, 0.23, 0.41, 0.83, 0.58. Assume that all eleven possible purchases form an independent class.
- Determine the distribution for \(X\), the amount purchased by Margaret.
- Determine the distribution for \(Y\), the amount purchased by Anne.
- Determine the distribution for \(Z = X + Y\), the total amount the two purchase.
Suggestion for part (c). Let MATLAB perform the calculations.
- Answer
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[r,s] = ndgrid(X,Y); [t,u] = ndgrid(PX,PY); z = r + s; pz = t.*u; [Z,PZ] = csort(z,pz);
% file npr06_18.m cx = [5 17 21 8 15 0]; cy = [8 15 12 18 15 12 0]; pmx = minprob(0.01*[37 22 38 81 63]); pmy = minprob(0.01*[77 52 23 41 83 58]); npr06_18 [X,PX] = canonicf(cx,pmx); [Y,PY] = canonicf(cy,pmy); [r,s] = ndgrid(X,Y); [t,u] = ndgrid(PX,PY); z = r + s; pz = t.*u; [Z,PZ] = csort(z,pz); a = length(Z) a = 125 % 125 different values plot(Z,cumsum(PZ)) % See figure Plotting details omitted