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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.
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.
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.
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.
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.
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.
Getting a 7 on Two Dice• There are 6 different ways of getting two dice to sum to 7: die A die B prob• In each case, the probability of 1 6 1/36 getting the required number on a single die is 1/6. 2 5 1/36• To get both numbers (so they add to 7), the probability uses 3 4 1/36 the AND rule: 1/6 x 1/6 = 1/36.• To sum up the 6 possibilities, 4 3 1/36 use the OR rule: only 1 of the 6 events can occur, but you 5 2 1/36 don’t care which one.• 6/36 = 1/6 6 1 1/36 total 6/36
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.
Sex Ratio in a Family of 3• Assume that the probability of a boy = 1/2 and the probability child child child of a girl = 1/2. #1 #2 #3• Enumerate each child separately for each of the 8 B B B possible families.• Each family has a probability B B G of 1/8 of occurring ( 1/2 x 1/2 x B G B 1/2).• Chance of 2 boys + 1 girl. B G G There are 3 families in which this occurs: BBG, BGB, and G B B GBB. Thus, the chance is 1/8 + 1/8 + 1/8 = 3/8. G B G G G B G G G
Different Probabilities for Different Phenotypes• Once again, a family of 3 child #1 child #2 child #3 total children, but this time the prob parents are heterozygous for Tay- T (3/4) T (3/4) T (3/4) 27/64 Sachs, a recessive T (3/4) T (3/4) t (1/4) 9/64 genetic disease. Each child thus has a 3/4 T (3/4) t (1/4) T (3/4) 9/64 chance of being normal T (3/4) t (1/4) t (1/4) 3/64 (TT or Tt) and a 1/4 chance of having the disease (tt). t (1/4) T (3/4) T (3/4) 9/64• Now, the chances for t (1/4) t (1/4) T (3/4) 3/64 different kinds of families is different. t (1/4) T (3/4) t (1/4) 3/64• chance of all 3 normal = t (1/4) t (1/4) t (1/4) 1/64 27/64. Chance of all 3 with disease = 1/64.
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.
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 ( p + q) = 1 n 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.
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. p + 2 pq + q 2 2
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.p + 3 p q + 3 pq + q = 1 3 2 2 3
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.
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.
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
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) = 15/1024 + 90/1024 = 105/1024.• 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.