WorkSHEET 9 - Jacaranda



WorkSHEET 9.1 Probability distributions Name: _________________________

| |Which of the following are discrete random variables? |Height is a continuous. | |

| |The heights of students in a Year 12 class. |Weight, to the nearest kg, is a discrete random variable. | |

| |The weights, to the nearest kg, of students in a Year 12 class. |Not a random variable, since match has already occurred. | |

| |The number of runs scored in a cricket test match in Brisbane in 2002.|Although infinite, still a discrete random variable. | |

| |The number of consecutive heads obtained when repeatedly tossing a |Discrete, since price is always quoted to the nearest $0.001 | |

| |coin. |Varies continuously, even when ‘full’ due to continuous pressure and | |

| |The price, per litre of petrol, in randomly selected stations in |temperature variation. | |

| |Queensland. | | |

| |The actual volume of petrol in the 1000 litre ‘unleaded petrol’ tanks | | |

| |at those same stations after being filled by a tanker. | | |

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| |Five coins are tossed simultaneously and the number of heads recorded.|Pr(5 heads) = Pr(0 heads) = [pic] = [pic] | |

| |Tabulate the probability distribution for the number of heads. |There are 5 ways to get 4 heads or 1 head. | |

| |Draw a probability distribution graph of the outcomes. |There are 10 ways to get 3 heads or 2 heads. | |

| | |[pic] | |

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| | |[pic] | |

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| |A die is ‘fixed’ so that certain numbers will appear more often. The |Since the sum of the probabilities must |

| |probability that a 6 appears is twice the probability of a 5 and |be 1, |

| |3 times the probability of a 4. The probabilities of 3, 2 and 1 are |[pic]+ [pic]+ [pic]+ [pic]+ [pic]+ x = 1 |

| |unchanged from a normal die. | |

| |The probability distribution table is given below. |Putting over a common denominator, |

| | |[pic] = 1 |

| |[pic] | |

| | |Collect like terms and remove fraction, |

| |Find: |3 + 11x = 6 |

| |The value of x in the probability distribution and hence complete the |x = [pic] |

| |probability distribution. | |

| |The probability of getting a ‘double’ with two of these dice. Compare |Note Pr(5) = [pic], Pr(4) = [pic] = [pic] |

| |with the ‘normal’ probability of getting a double. | |

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| | |[pic] |

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| | |Probability of a ‘double’ is given by |

| | |Pr(1) ( Pr(1) + Pr(2) ( Pr(2) + … +Pr(6) |

| | |( Pr(6) |

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| | |Pr(double) = [pic]( [pic]( [pic]( [pic]+[pic]( [pic]+ |

| | |[pic]( [pic]+ [pic]( [pic]+ [pic]( [pic] |

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| | |Convert each product to a decimal. |

| | |Pr(double) = 0.02777 + 0.02777 + 0.02777 |

| | |+ 0.00826 + 0.01860 + 0.07438 |

| | |Pr(double) = 0.1846 |

| | |The ‘normal’ probability of a double is 0.1666. |

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| |Show that p(x) = [pic], for x = 1,2, … 6 |Set up a table of probabilities | |

| |is a probability distribution. | | |

| | |[pic] | |

| |State Pr (2 < x < 5) | | |

| | |Calculate the sum of the probabilities. | |

| | |Sum = [pic]= 1 | |

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| | |Pr (2 < x < 5) = [pic]+ [pic]+ [pic] | |

| | |= [pic] | |

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| |Find the expected value of the following discrete probability |Use formula E(X) = ΣxPr(X = x) |

| |distribution. |E(X) = 1(0.1) + 2(0.15) + 3(0.25) |

| |[pic] |+ 4(0.05) + 5(0.45) |

| | |E(X) = 0.1 + 0.3 + 0.75 + 0.2 + 2.25 |

| | |= 3.6 |

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| |Consider the following gambling game, based on the outcome of the |Set up the probability distribution table. | |

| |total of 2 dice: |[pic] | |

| |– if the total is a perfect square, you win $4 | | |

| |– if the total is 2, 6, 8 or 10, you win $1 |Add a row, which indicates win (+) or loss(–). | |

| |– otherwise, you lose $2. |[pic] | |

| | |Use formula E(gain) = ΣGain(x) Pr(X = x) | |

| |Find the expected value of this game. |E(gain) = 1([pic]) –2([pic]) + 4([pic]) –2([pic]) | |

| | |+ 1([pic]) –2([pic]) +1([pic]) + 4([pic]) | |

| |Determine if it is a fair game. |+ 1([pic]) –2([pic]) –2([pic]) | |

| | |E(Gain) = [pic] | |

| | |[pic] | |

| | |E(Gain) = [pic]– [pic]= [pic] | |

| | |E(Gain) = 0.333 | |

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| | |This game is ‘unfair’ — you stand to gain about $0.33 every time you | |

| | |play! | |

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| |Find the missing profit (or loss) so that the following probability |Let y = profit/loss for x = 10. |

| |table has an expected value of 0. |E(Gain) = ΣGain(x) Pr(X = x) |

| |[pic] |E(Gain) = –3(0.1) + 4(0.06) – 2(0.25) + 5(0.16) –8(0.09) + 12(0.21) + |

| | |y(0.13) |

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| | |Simplify and set E(gain) = 0 |

| | |–0.3 + 0.24 – 0.5 + 0.8 – 0.72 + 2.52 + 0.13y = 0 |

| | |2.04 + 0.13y = 0 |

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| | |Solve for y |

| | |[pic] |

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| |For the following probability distribution calculate: |Using a table of values (or the Maths Quest spreadsheet ‘Prob | |

| |E(X) |distribution’): | |

| |E(2X) | | |

| |E(X + 2) |E(X) ( 1.22 | |

| |E(X2) |E(2X) ( 2.44 | |

| |E(X2) – [E(X)]2. |E(X + 2) ( 3.22 | |

| | |E(X2) ( 5.24 | |

| |[pic] |E(X2) – [E(X)]2 ( 3.7516 | |

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| |Three players play the following game for a prize pool of $210. |Because, in theory, this game could go on forever, determine (relative) |

| | |probabilities as follows. |

| |Alice tosses a coin — if it is heads she wins. If not, then Betty | |

| |tosses the coin — if it is heads she wins. If not, then Carla tosses |In round 1, |

| |the coin — if it is heads she wins. If not, then Alice tosses the coin|Alice has a chance of winning, while Betty has a |

| |again, winning if it is a head … and so on. |× chance, and Carla has a × × chance. |

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| |Find the expected value of each person in this game. |In Round 2, |

| | |Alice has a × × × chance … and so on. |

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| | |These probabilities are tabulated below. |

| | |[pic] |

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| | |By looking at each row, the probabilities are in the ratio of 4 : 2 : 1 |

| | |Thus Alice has 4 ‘chances’, Betty has 2 and Carla has 1. |

| | |E(Alice) = (210) = $120 |

| | |E(Betty) = (210) = $60 |

| | |E(Carla) = (210) = $30 |

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