There are 39 students in a psychology class. The class has been divided into 10 groups; 9 groups have 4 students in them, 1 group only has 3 students. The instructor puts all of the students names in a hat and then randomly draws 1 name. What are the chances that the one name will be the name of someone in the small, 3-person group?

Answer: Our required probability is 0.1875.Step-by-step explanation:Since we have given that Number of students = 39Total Number of groups = 10Number of groups containing 4 students = 9Number of groups containing 3 students = 1So, Probability of getting group of 4 students = [tex]\dfrac{9}{10}[/tex]Probability of getting group of 3 students = [tex]\dfrac{1}{10}[/tex]Using the "Bayes theorem":Probability that the one name will be the name of someone in the small 3-person group is given by[tex]\dfrac{\dfrac{1}{10}\times \dfrac{3}{39}}{\dfrac{1}{10}\times \dfrac{3}{39}+\dfrac{9}{10}\times \dfrac{4}{39}}\\\\=\dfrac{\dfrac{3}{390}}{\dfrac{3+13}{390}}\\\\=\dfrac{3}{16}\\\\=0.1875[/tex]Hence, our required probability is 0.1875.