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ASU MAT 142 - Probability

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VocabularyIn order to discuss probability we will need a fair bit of vocabulary.Probability is a measurement of how likely it is for something to happen.An experiment is a process that yields an observation.Example 1. Experiment 1. The experiment is to toss a coin and observe whether it landsheads or tails.Experiment 2. The experiment is to toss a die and observe the number of spots.An outcome is the result of an experiment.Example 2. Experiment 1. One possible outcome is heads, which we will designate H.Experiment 2. One possible outcome is 5.Sample space is the set of all possible outcomes, we will usually represent this set with S.Example 3. Experiment 1. S = {H, T }Experiment 2. S = {1, 2, 3, 4, 5, 6}.An event is any subset of the sample space. Events are frequently designated as E or E1,E2, etc. if there are more than one.Example 4. Experiment 1. One possible event is landing heads up. E = {H}.Experiment 2. One event might be getting an even number, a second event might be gettinga number greater than or equal to 5. E1= {2, 4, 6}, E2= {5, 6}.A certain event is guaranteed to happen, a “sure thing”. An impossible event is onethat cannot happen.Example 5. From our Experiment 1. If the event E1is landing heads up or landing tailsup, then E1= {H, T } and we can see that E1is a certain event. If the event E2is getting a6, then E2= {} and we can see than E2is impossible.Outcomes in a sample space are said to be equally likely if every outcome in the samplespace has the same chance of happening, i.e. one outcome is no more likely than any other.Basic Computation of a ProbabilityIf E is an event in an equally likely sample space, S, then the probability of E, denoted p(E)is computedp(E) =n(E)n(S)Example 6. Experiment 1. E1= {H, T }, so p(E1) =n(E1)n(S)=22= 1.Experiment 2. E2= {5, 6}, so p(E2) =n(E2)n(S)=26=13.12 ProbabilityYou should note that since n(∅) = 0, p(∅) = 0 and sincen(S)n(S)= 1, p(S) = 1.Relative frequency is the number of times a particular outcome occurs dvided by thenumber of times the experiment is performed.Example 7. Suppose you toss a fair coin 10 times and observe the results: heads occurred6 times and tails occurred 4 times. In this case the relative frequency of heads would be610= 0.6.Now suppose you toss a fair coin 1000 times and observe the results: heads occurred 528times and tails occured 472 times. Now the relative frequency of heads is5281000= 0.528.Something which can be observed from the previous two experiments is called the Law ofLarge Numbers. The Law of Large Numbers: If an experiment is repeated a largenumber of times, the relative frequency tends to get closer to the probability.Example 8. In the previous example, we observed the outcome heads. This was a fair coinso we know p(H) = 0.5. Notice that when we performed the experiment 1000 times, therelative frequency was closer to 0.5 than when we performed the experiment 10 times.OddsThe odds for an event, E, with equally likely outcomes are:o(E) = n(E) : n(E0)and the odds against the event E are:o(E0) = n(E0) : n(E). Note, these are read as a to b.Example 9. Toss a fair coing twice and observe the outcome. The sample space isS = {HH, HT, T H, T T }.If E is getting two tails, theno(E) = 1 : 3,the odds for E are 1 to 3. Alsoo(E0) = 3 : 1,the odds against E are 3 to 1.Some ApplicationsProbability is used in genetics. When Mendel experimented with pea plants he discoveredthat some genes were dominant and others were recessive. This means that given one genefrom each parent, the dominant trait will show up unless a recessive gene is received fromeach parent. This is often demonstrated by using a “Punnett square”. Here is a typicalPunnett square:R Rw wR wRw wR wRProbability 3The letters along the top of the table represent the gene contribution from one parent andthe letters down the left-hand side of the table represent the gene contribution from theother parent. Each cell of the table contains a genetic combination for a possible offspring.This particular Punnett square represent the crossing of a pure red flower pea with a purewhite flower pea. The offspring will each inherit one of each gene; since red is dominanthere, all offspring will be red.Example 10. Suppose we cross a pure red flower pea plant with one of the offspring thathas one of each gene. The resulting Punnett square would beR Rw wR wRR RR RRFrom this we can see that the probability of producing a pure red flower pea is24=12, andwe can see that each offspring would produce red flowers.Example 11. This time we will cross two of the offspring which have one of each gene. ThePunnett square isw Rw ww wRR Rw RRHere we can find that the probability of an offspring producing white flowers is14.Example 12. Sickle cell anemia is inherited. This is a co-dominant disease. A person withtwo sickle cell genes will have the disease while a person with one sickle cell gene will havea mild anemia called sickle cell trait. Suppose a healthy parent (no sickle cell gene) and aparent with sickle cell trait have children. Use a Punnett square to determine the followingprobabilities.(1) the child has sickle cell(2) the child has sickle cell trait(3) the child is healthySolution: We will use S for the sickle cell gene and N for no sickle cell gene. Our Punnettsquare isN NS SN SNN NN NN(1) No offspring have two sickle cell genes so this probability is 0.(2) We see that two of the offspring will have one of each gene, so this probability is24=12.(3) Two of the children will have no sickle cell genes, this probability is24=12.4 ProbabilityBasic Properties of ProbabilityWe will need to start with one more new term. Two events are mutually exclusive if theycannot both happen at the same time, i.e. E ∩ F = ∅. So, if p(A ∩ B) = 0 then A and Bare mutually exclusive,We have the following rules for probability:Basic Probability Rulesp(∅) = 0,p(S) = 1,0 ≤ p(E) ≤ 1Example 13. Roll a die and observe the number of dots; S = {1, 2, 3, 4, 5, 6}. Given E isthe event that you roll a 15, F is the event that you roll a number between 1 and 6 inclusive,and G is the event that you roll a 3. Find p(E), p(F ), and p(G).Solution: Since 15 is not in the sample space E = ∅ and p(E) = 0. F = S, so p(F ) = 1.Lastly, G = {3}, so p(G) =n(G)n(S)16.Example 14. Roll a pair of dice, one red and one white. Find the probabilities:(1) the sum of the pair is 7(2) the sum is greater than 9(3) the sum is not greater than 9(4) the sum is greater than 9 and even(5) the sum is


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ASU MAT 142 - Probability

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