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When the occurrence of one event is independent of the occurrence of other events, the product rule is used: The probability that two independent events will both occur is the product of their separate probabilities. This is known as the and rule. For example, the probability of throwing a die two times and getting a four and then a six, in that order, is P 1/6 1/6 1/36
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Thus, for unordered events, we can obtain the probability by combining rules 1 and 2. The binomial theorem (rule 3) provides the shorthand method. To use rule 3, we must state the theorem as follows: If the probability of an event (X) is p and an alternative (Y ) is q, then the probability in n trials that event X will occur s times and Y will occur t times is P n! s t pq s!t!
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3. Binomial Theorem
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The binomial theorem is used for unordered events: The probability that some arrangement will occur in which the nal order is not speci ed is de ned by the binomial theorem. For example, what is the probability when tossing two pennies simultaneously of getting a head and a tail We will look more closely at how to use the rules of probability to answer this question.
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In this equation, s t n, and p q 1. The symbol !, as in n!, is called factorial, as in n factorial, and is the product of all integers from n down to one. For example, 7! 7 6 5 4 3 2 1. Zero factorial equals one, as does anything to the power of zero (0! n0 1). Now, what is the probability of tossing two pennies at once and getting one head and one tail In this case, n 2, s and t 1, and p and q 1/2. Thus, P 2! (1/2)1(1/2)1 1!1! 2(1/2)2 1/2
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This is, of course, our original answer. Now on to a few more genetically relevant problems. What is the probability that a family with six children will have ve girls and one boy (We assume that the probability of either a
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II. Mendelism and the Chromosomal Theory
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4. Probability and Statistics
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Use of Rules
son or a daughter equals 1/2.) Since the order is not speci ed, we use rule 3: P 6! (1/2)5(1/2)1 5!1! 6(1/2)6 6/64 3/32
binomial expansion formula. First, if you have dif culty calculating the term, you can use Pascal s triangle to get the coef cients: 1 1 1 1 1 1 5 4 10 3 6 10 2 3 4 5 1 1 1 1 1
What would happen if we asked for a speci c family order, in which four girls were born, then one boy, and then one girl This would entail rule 2; for a sequence of six independent events: P 1/2 1/2 1/2 1/2 1/2 1/2 1/64
When no order is speci ed, the probability is six times larger than when the order is speci ed; the reason is that there are six ways of getting ve girls and one boy, and the sequence 4-1-1 is only one of them. Rule 3 tells us that there are six ways. These are (letting B stand for boy and G for girl) as follows:
Birth Order 1 B G G G G G 2 G B G G G G 3 G G B G G G 4 G G G B G G 5 G G G G B G 6 G G G G G B
Pascal s triangle is a triangular array made up of coef cients in the binomial expansion. It is calculated by starting any row with a 1, proceeding by adding two adjacent terms from the row above, and then ending with a 1. For example, the next row would be 1, (1 5), (5 10), (10 10), (10 5), (5 or 1, 6, 15, 20, 15, 6, 1 1), 1
Let us look at yet another problem. If two persons, heterozygous for albinism (a recessive condition), have four children, what is the probability that all four children will be normal The answer is simply (3/4)4 by rule 2. What is the probability that three will be normal and one albino If we specify which of the four children will be albino (e.g., the fourth), then the probability is (3/4)3(1/4)1 27/256. If, however, we do not specify order, P 4! (3/4)3(1/4)1 4(3/4)3(1/4)1 3!1! 4(27/256) 108/256
These numbers give us the combinations for any psqt term. That is, in our previous example, n 4; so we use the (n 1), or fth, row of Pascal s triangle. (The second number in any row of the triangle gives the power of the expansion, or n. Here, 4 is the second number in the row.) We were interested in the case of one albino child in a family of four children, or p3q1, where p is the probability of the normal child (3/4) and q is the probability of an albino child (1/4). Hence, we are interested in the (t 1) that is, the (1 1) or the second term of the fth row of Pascal s triangle, which will tell us the number of ways of getting a four-child family with one albino child. That number is 4. Thus, using Pascal s triangle, we see that the solution to the problem is 4(3/4)3(1/4)1 108/256
This is precisely four times the ordered probability because the albino child could have been born rst, second, third, or last. The formula for rule 3 is the formula for the terms of the binomial expansion. That is, if ( p q)n is expanded (multiplied out), the formula (n!/s!t!)psqt gives the probability for any one of these terms, given that p q 1 and that s t n. Since there are (n 1) terms in the binomial, the formula gives the probability for the term numbered (t 1). Two bits of useful information come from recalling that rule 3 is in reality the
This is the same as the answer we obtained the conventional way. The second advantage from knowing that rule 3 is the binomial expansion formula is that we can now generalize to more than two outcomes. The general form for the multinomial expansion is ( p q r )n and the general formula for the probability is P n! psqtru s!t!u!
where s t u n and p q r 1 For example, our albino-carrying heterozygous parents may want an answer to the following question: If we have ve children, what is the probability that we will have two normal sons, two normal daughters, and one albino son (This family will have no albino daughters.) By rule 2, the probability of
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