Probability II: Binomial Expansion v2.1/05. Introduction. How to do a problem. Prerequisite: Probability I & Sex Inheritance

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1 Probability II: Binomial Expansion v2.1/05 Prerequisite: Probability I & Sex Inheritance Name Introduction In the Probability I topic, you considered probability of a single event. You learned how to compute the probability of parents whose genotypes were Aa x Aa producing a child whose genotype was aa. But what is the probability of those parents producing 3 children containing the dominant genotype AA or Aa and 2 children of genotype aa? This is more complicated. To answer questions that refer to the probability of multiple events, we still use the Product Rule, but we need to use a little more math. As biologists, we will be using this math as a tool, so we will not explain to you why these formulas work. (Your math teachers can do this if you like!) How to do a problem Problem #1: For a family of three children, what is the probability of having one girl and two boys? Solution: 1. First you must do the basic genetics. This includes working out the gametes, and drawing a Punnett Square. So: Man's genotype: Woman's genotype: XX Man's gametes: X and Y Woman's gametes: X and X X Y 2. Then consider the probability of each separate event: a = chance of having a boy = 2 / 4 = 1 / 2 b = chance of having a girl = 2 / 4 = 1 / 2 3. Then put the separate probabilities into the following formula: (a + b) n where a = chance of one event (having a boy) b = chance of other event (having a girl) n = number of total events (for a family of 3 children, n = 3) The term (a + b) is called a binomial. 4. So for this problem, we would write: (a + b) 3 5. Now expand the binomial; in other words, multiply the expression (a + b) times itself three times. Fortunately, we can give you a short cut for doing this. (continued next page) Probability II: Binomial Expansion page 1

2 Expanding a binomial: A. (a + b) 3 = a 3 + a 2 b 1 + a l b 2 + b Note that we have first raised "a" to the highest power. 2. For the next term, we bring down the "a" by one power, to 2. And we begin to include the b starting at a power of Here the "a is brought down again, by one power, and "b" is raised to the 2nd power. 4. Finally a is down to zero power, and since any number raised to the power of zero equals 1, we do not have to include it. "b" is raised by one to the third power. 5. In each term the exponent total number is 3: 3 + 0; 2 + 1; 1 + 2; and B. Secondly, you must fill in the coefficients for each term of the expanded binomial. (a + b) 3 =? a 3 +? a 2 b 1 +? a l b 2 +? b 3 To do this we call on Blaise Pascal, a 17th century mathematical genius who figured out a clever game. We call it Pascal's Triangle. Here it is: Do you see a pattern? Pascal's Triangle can be built on indefinitely. To build the triangle: Start by writing Now on the next line, write a 2 1 And finally, finish with a 1 Then add the two numbers above and place the sum in the middle. For the third line of the Pascal's Triangle, you would continue in the same manner: Start by writing a And finish with a 1 Obtain each of these numbers by adding separately the two numbers just above. Probability II: Binomial Expansion page 2

3 Can you write the fifth line on this Pascal s Triangle? C. Now let us return to the poor little incompletely expanded binomial: (a + b) 3 = a 3 + a 2 b 1 + a l b 2 + b 3 Answer: 1, 5, 10, 10, 5, 1 To find the coefficients for the power of three, refer to the third line of your Pascal's Triangle. (If it were raised to the power of 5, you would use fifth line. The power that it is raised to indicates the line number of Pascal's Triangle that gives the correct coefficients.) Place the numbers, in order, to the left of each term. (a + b) 3 = 1 a a 2 b a l b b 3 This is the completely expanded binomial (a+b) 3. Back to solving the problem 6. Now choose the term from this expanded binomial that represents the combination of events you have been studying. The question was: For a family of three children, what is the probability of having one girl and two boys? a = chance of having a boy = 1 / 2 b = chance of having a girl = 1 / 2 From the following expanded binomial : a a 2 b + 3 ab 2 + b 3, two boys and one girl is represented by the term 3 a 2 b. 7. Substitute in the values for a and b in the term you have chosen: a = 1 / 2 3 a 2 b = 3 ( 1 / 2 ) 2 ( 1 / 2 ) b = 1 / 2 = 3 ( 1 / 4 ) ( 1 / 2 ) ANSWER = 3 / 8 8. The answer 3 / 8, tells us that for families of three children, the probability of having one girl and 2 boys is 3 / 8. OR, for every eight families of three children, three families will have one gir1 and two boys. Probability II: Binomial Expansion page 3

4 Problem #2 (since the previous problem was so spread out, we will go through another problem with little explanation) Question: In a family of five children, what would the probability be of two brown-eyed heterozygotes having four brown-eyed children and one blue-eyed child? (Assume brown eyes is dominant to blue) Solution: Step 1: Genetics Key: B=brown eyes, b=blue eyes Parents: Bb x Bb Gametes: B, b x B, b B B BB b Bb b Bb bb Step 2: Probability of separate events a = chance of brown eyes = 3 / 4 b = chance of blue eyes = 1 / 4 Step 3: Expand binomial: (a + b) 5 remember 5 children a. (a + b) 5 = a 5 + a 4 b 1 + a 3 b 2 + a 2 b 3 + a 1 b 4 + b 5 b. Find coefficients by Pascal's Triangle and put them in the expanded binomial a a 4 b a 3 b a 2 b a 1 b b 5 c. Choose the term that represents four brown-eyed children and one blue-eyed child: 5 a 4 b d. Substitute: a = 3 / 4 5 a 4 b = 5 ( 3 / 4 ) 4 ( 1 / 4 ) b = 1 / 4 = 5 ( 81 / 256 ) ( 1 / 4 ) ANSWER = 405 / 1024 Note: When doing probability with experimental data, the final fraction is not usually reduced. If the probability had been 256 / 1024, it would be a different answer to reduce it to 1 / 4. (You lose a sense of the sample size.) Probability II: Binomial Expansion page 4

5 Try these sample problems. If you have trouble, refer to the problems above. When you think you have completed them correctly, check the answer sheet on back. A. For a family of 4 children, what is the probability of having 2 girls and 2 boys? B. Two right-handed people marry and have 6 children. Three of these children are left handed, 3 of these children are right handed. What is the probability of this? (Assume right-handedness is dominant to left-handedness) Probability II: Binomial Expansion page 5

6 Probability II Answer Sheet A. Four children; chance of having 2 girls and 2 boys: 1. X Y a = chance of having a boy, 1/2 b = chance of having a girl, 1/2 2. (a + b) 4 = a 4 + a 3 b 1 + a 2 b 2 + a 1 b 3 + b 4 3. Pascal's Triangle 1 2 a a 3 b + 6 a 2 b ab b a 2 b 2 represents 2 boys and 2 girls. Substitute: a = 1/2 6 a 2 b 2 = 6 ( 1 / 2 ) 2 ( 1 / 2 ) 2 b = 1/2 = 6 ( 1 / 4 ) ( 1 / 4 ) = 6 / 16 B. Six children born to heterozygote parents, 3 lefties and 3 righties. 1. Parents: Hh x Hh H h H HH Hh h Hh hh a = chance of being right handed = 3/4 b = chance of being left handed = 1/4 2. (a + b) 6 = a 6 + a 5 b 1 + a 4 b 2 + a 3 b 3 + a 2 b 4 + a 1 b 5 + b 6 3. Pascal s Triangle: = a a 5 b a 4 b a 3 b a 2 b a 1 b 5 + b lefties and 3 righties = 20 a 3 b 3. Substitute: a = 3/4 20 a 3 b 3 = 20 ( 3 / 4 ) 3 ( 1 / 4 ) 3 b =1/4 = 20 ( 27 / 64 ) ( 1 / 64 ) = 540 / 4096 Probability II: Binomial Expansion page 6