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Probability

Probability. Probability: what is the chance that a given event will occur? For us, what is the chance that a child, or a family of children, will have a given phenotype?

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Probability

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  1. Probability • Probability: what is the chance that a given event will occur? For us, what is the chance that a child, or a family of children, will have a given phenotype? • Probability is expressed in numbers between 0 and 1. Probability = 0 means the event never happens; probability = 1 means it always happens. • The total probability of all possible event always sums to 1.

  2. Definition of Probability • The probability of an event equals the number of times it happens divided by the number of opportunities. • These numbers can be determined by experiment or by knowledge of the system. • For instance, rolling a die (singular of dice). The chance of rolling a 2 is 1/6, because there is a 2 on one face and a total of 6 faces. So, assuming the die is balanced, a 2 will come up 1 time in 6. • It is also possible to determine probability by experiment: if the die were unbalanced (loaded = cheating), you could roll it hundreds or thousands of times to get the actual probability of getting a 2. For a fair die, the experimentally determined number should be quite close to 1/6, especially with many rolls.

  3. The AND Rule of Probability • The probability of 2 independent events both happening is the product of their individual probabilities. • Called the AND rule because “this event happens AND that event happens”. • For example, what is the probability of rolling a 2 on one die and a 2 on a second die? For each event, the probability is 1/6, so the probability of both happening is 1/6 x 1/6 = 1/36. • Note that the events have to be independent: they can’t affect each other’s probability of occurring. An example of non-independence: you have a hat with a red ball and a green ball in it. The probability of drawing out the red ball is 1/2, same as the chance of drawing a green ball. However, once you draw the red ball out, the chance of getting another red ball is 0 and the chance of a green ball is 1.

  4. The OR Rule of Probability • The probability that either one of 2 different events will occur is the sum of their separate probabilities. • For example, the chance of rolling either a 2 or a 3 on a die is 1/6 + 1/6 = 1/3.

  5. NOT Rule • The chance of an event not happening is 1 minus the chance of it happening. • For example, the chance of not getting a 2 on a die is 1 - 1/6 = 5/6. • This rule can be very useful. Sometimes complicated problems are greatly simplified by examining them backwards.

  6. Combining the Rules • More complicated situations involve combining the AND and OR rules. • It is very important to keep track of the individuals involved and not allow them to be confused. This is the source of most people’s problems with probability. • What is the chance of rolling 2 dice and getting a 2 and a 5? The trick is, there are 2 ways to accomplish this: a 2 on die A and a 5 on die B, or a 5 on die A and a 2 on die B. Each possibility has a 1/36 chance of occurring, and you want either one or the other of the 2 events, so the final probabilty is 1/36 + 1/36 = 2/36 = 1/18.

  7. Getting a 7 on Two Dice • There are 6 different ways of getting two dice to sum to 7: • In each case, the probability of getting the required number on a single die is 1/6. • To get both numbers (so they add to 7), the probability uses the AND rule: 1/6 x 1/6 = 1/36. • To sum up the 6 possibilities, use the OR rule: only 1 of the 6 events can occur, but you don’t care which one. • 6/36 = 1/6

  8. Probability and Genetics • The probability that any individual child has a certain genotype is calculated using Punnett squares. • We are interested in calculating the probability of a given distribution of phenotypes in a family of children. • This is calculated using the rules of probability.

  9. Sex Ratio in a Family of 3 • Assume that the probability of a boy = 1/2 and the probability of a girl = 1/2. • Enumerate each child separately for each of the 8 possible families. • Each family has a probability of 1/8 of occurring ( 1/2 x 1/2 x 1/2). • Chance of 2 boys + 1 girl. There are 3 families in which this occurs: BBG, BGB, and GBB. Thus, the chance is 1/8 + 1/8 + 1/8 = 3/8.

  10. Different Probabilities for Different Phenotypes • Once again, a family of 3 children, but this time the parents are heterozygous for Tay-Sachs, a recessive genetic disease. Each child thus has a 3/4 chance of being normal (TT or Tt) and a 1/4 chance of having the disease (tt). • Now, the chances for different kinds of families is different. • chance of all 3 normal = 27/64. Chance of all 3 with disease = 1/64.

  11. Different Probabilities for Different Phenotypes, p. 2 • Chance of 2 normal + 1 with disease: there are 3 families of this type, each with probability 9/64. So, 9/64 + 9/64 + 9/64 = 27/64. • Chance of “at least” one normal child. This means 1 normal or 2 normal or 3 normal. Need to figure each part separately, then add them. --1 normal + 2 diseased = 3/64 + 3/64 + 3/64 = 9/64. --2 normal + 1 diseased = 27/64 (see above) -- 3 normal = 27/64 --Sum = 9/64 + 27/64 + 27/64 = 63/64. • This could also be done with the NOT rule: “at least 1 normal” is the same as “NOT all 3 diseased”. The chance of all 3 diseased is 1/64, so the chance at least 1 normal is 1 - 1/64 = 63/64.

  12. Larger Families: Binomial Distribution • The basic method of examining all possible families and counting the ones of the proper type gets unwieldy with big families. • The binomial distribution is a shortcut method based on the expansion of the equation to the right, where p = probability of one event (say, a normal child), and q = probability of the alternative event 9mutant child). n is the number of children in the family. • Since 1 raised to any power (multiplied by itself) is always equal to 1, this equation describes the probability of any size family.

  13. Binomial for a Family of 2 • The expansion of the binomial for n = 2 is shown. The 3 terms represent the 3 different kinds of families: p2 is families with 2 normal children, 2pq is the families with 1 normal and 1 mutant child, and q2 is the families with 2 mutant children. • The coefficients in front of these terms: 1, 2, and 1, are the number of different families of the given type. Thus there are 2 different families with 1 normal plus 1 mutant child: normal born first and mutant born second, or mutant born first and normal born second. • As before, p = 3/4 and q = 1/4. • Chance of 2 normal children = p2 = (3/4)2 = 9/16. • Chance of 1 normal plus 1 mutant = 2pq = 2 * 3/4 * 1/4 = 6/16 = 3/8.

  14. Binomial for a Family of 3 • Here, p3 is a family of 3 normal children, 3p2q is 2 normal plus 1 affected, 3pq2 is 1 normal plus 2 affected, and q3 is 3 affected. • The exponents on the p and q represent the number of children of each type. • The coefficients are the number of families of that type. • Chance of 2 normal + 1 affected is described by the term 3p2q. Thus, 3 * (3/4)2 * 1/4 = 27/64. Same as we got by enumerating the families in a list.

  15. Larger Families • To write the terms of the binomial expansion for larger families, you need to get the exponents and the coefficients. • Exponents are easy: you just systematically vary the exponents on p and q so they always add to n. Start with pnq0 (remembering that anything to the 0 power = 1), do pn-1q1, then pn-2q2, etc. • Coefficients require a bit more work. There are several methods for finding them. I am going to show you Pascal’s Triangle, but other methods are also commonly used.

  16. Pascal’s Triangle • Is a way of finding the coefficients for the binomial in a simple way. • Start by writing the coefficients for n = 1: 1 1. • Below this, the coefficients for n = 2 are found by putting 1’s on the outside and adding up adjacent coefficients from the line above: 1, 1 + 1 = 2, 1. • Next line goes the same way: write 1’s on the outsides, then add up adjacent coefficients from the line above: 1, 1+2 = 3, 2+1 = 3, 1. • For n = 5, coefficients are 1, 5, 10, 10, 5, 1.

  17. More Pascal’s Triangle • Now apply the coefficients to the terms. For n = 5, the terms with appropriate exponents are p5, p4q, p3q2, p2q3, pq4, and q5. • The coefficients are 1, 5, 10, 10, 5, 1. So the final equation is p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 = 1. • Using this: what is the chance of 3 normal plus 2 affected children? The relevant term is 10p3q2. The exponents on p and q determine how many of each kind of child is involved. The coefficient, 10, says that there are 10 families of this type on the list of all possible families. • So, the chance of the desired family is 10p3q2 = 10 * (3/4)3 * (1/4)2 = 10 * 27/64 * 1/16 = 270/1024

  18. More with this Example • What is the chance of 1 normal plus 4 affected? The relevant term is 5pq4. So, the chance is 5 * (3/4) * (1/4)4 = 5 * 3/4 * 1/256 = 15/1024. • What is the chance of 1 normal or 2 normal? Sum of the probabilities for 5pq4 (1 normal) and 10p2q3 (2 normal) = 90/1024 + 15/1024 = 285/1024. • What is the chance of at least 4 normal? This means 4 normal or 5 normal. Add them up. • What is the chance of at least 1 normal? Easiest to do with the NOT rule: 1 - chance of all affected. • What is the chance that child #3 is normal? Trick question. For any individual child, the probability is always the simple probability from the Punnett square: 3/4 in this case.

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