Suppose that there are six prospective jurors, four men (M) and two women (W), who might be impaneled to sit on the jury in a criminal case. Two jurors are randomly selected from these six to fill the two remaining jury seats. (a) List the simple events in the experiment. (HINT: There are 15 simple events if you ignore the order of selection of the two jurors. Select all that apply.)

Answer:W1W2; W1M1; W1M2; W1M3; W1M4; W2M1; W2M2; W2M3; W2M4; M1M2; M1M3; M1M4; M2M3; M2M4; M3M4.Step-by-step explanation:Applying combinatorial analysis, the total number of simple events is given by:[tex]E= \frac{6!}{2!(6-2)!}=\frac{6*5*4!}{2*4!}\\E= \frac{6*5}{2} = 15[/tex]Let the four men be M1, M2 ,M3, M4 nad the two women W1 and W2All of the possible events with W1 on the jury are:W1W2; W1M1; W1M2; W1M3; W1M4;The remaining possible events with W2 on the jury are:W2M1; W2M2; W2M3; W2M4.The remaining possible events with M1 on the jury are:M1M2; M1M3; M1M4The remaining possible events with M2 on the jury are:M2M3; M2M4The last possible event is:M3M4