A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?	Pick:  A) 1/10  B) 1/5  C) 3/10  D) 2/5  E) 1/2
We have three units - Tom (T), Mary (M) &amp; the Rest (R).   Total ways of making a community of 3 from 6 people = 6 x 5 x 4 = 120   Now ways the community can be formed with Tom in it and Mary NOT in it:   TRR   RTR   RRT   TRR posb = 1 (way of picking Tom) x 4 (4 people left excluding Tom and Mary) x 3 (3 people left after excluding Tom, Mary and the person selected before) = 12   RTR and RRT have the possibility as TRR. Therefore possibilities the community can be formed with Tom in it and Mary NOT in it = 12 + 12 + 12= 
<gadget id="calculator">12 + 12 + 12</gadget>
<output>36</output>
36   Hence probability=36/120=3/10   Option 3/10
<result>C</result>
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?