Question

29 Principles of Probability first developed in Section I1-1. Finally, you saw that if a 1:1 ratio, similar to that for heads and tails on coins, is assumed for the binth of boys and girls in human populations, the same combinations would be expected for babies as for coins (compare Section I1-4 with Section II1-5) Expectations for various combinations in groups of given size (n) can be obtained mathemati- cally Mendel and others recognized that combinations can be calculated by expanding the binomial (ab)", in which n is the size of the group, a is the probability of the first event, b is the prob- ability of the atremative event and ab=1. In the example involving babies, a girls 1/2 and b babies are bom in the hospital in 1 day, expand (a + b) a probability of 1/2. Now, if you wish to consider the case in which four 4a'b + 6ab + 4ab +b probability of boys Expectation of 4 girls a-(1/2)* = 1/16 Expectation of 3 girls, 1 boy Expectation of 2 girls, 2 boys 6a'b6(1/2) ( 1/2)2-6/16 3/8 Expectation of 1 girl, 3 boys 4ab= 4(1/2) (1/2)' = 4/16 1/4 Expectation of 4 boys 4a'b 4(1/2)'(1/2) 4/16 1/4 b-(1/2) = 1/16 1. If five babies are born in a given hospital on the same day, what is the chance of (a) 5 boys; (b) 4 boys, 1 girl; (c) 3 boys, 2 girls; (d) 2 boys, 3 girls; (e) 1 boy, 4 girls; () 5 girls? To answer this question, you expand the binomial (a + b)- a +5a'b+ 10a'b substitute a 1/2 and b-1/2 in each term. Do these steps, and answer each part of the ques- tion in the following spaces. 10a2b +5ab+ b and + (b) (a) (d)

Answer 1

given 5 babies are born in hospital on the same day. an What is the chance of having: (a) s boys (6) 4boys igirl (d) 2 Bboys 3 girls (e) i boy 4 girls (c) z boys sgirls (t) s girls. Ans Now the expantion of the bioromial (a+b) is : where (a+b) = as + sahb + 10a3b²+ 10 2²6² + sab" + 65 a= Probability of a boy = 1/2 b= Probability 2% a girl = 1/2 Therefore: (a) Probability of 5 boys = a = (1/2) = 1/32 (6) Probability of 4boys and gin = sahb= 5(1/2) + (1/2) = 5/32 (c) Probability of 3 boys and a girls = 10a3b² = 10x 1/2) + (1/2)² - 10*/8 *'/4 = 10/32 = 5/16 (d) Probability of 2boys and 3 girls = 10a?b= 10 x (/2) (2) = 10 x 1/8 x 4 = 10/32 = 5/16 (1) Probability of Iboy and a girls = sab" = 5x (1/2) × (1/2) 4 = 5 x 32 = 5/32 tt) Probability of 5 girls = b = (1/2) = 1/32