total # of M&Ms 12 (3 green) 4. [7 pts] There are 12 M&M's in a bag, 3 of them are blue, 5 are red, and 4 of them are green. We are going to draw 2 of them from the bag, one at a time without replacing them. (Hint: it might help to draw a probability tree to help you with this question). Calculate the following probabilities (showing your work and giving your solutions as percentages with 2 decimal places): Drawing a red M&M and then a green M&M The probability of drawing a red M&M is 5/12. This leaves 11 M&Ms left with 4 of them being green, making the probability of drawing a green M&M 4/11. P(red on 1st draw)+P(green on 2nd draw) = 20 0.15ISIS =0.151515 100 15.15% Drawing both a blue and red M&M (their order doesn't matter). sample space = 12c event space = 3c, Sc. c + Sej 3c, co =(3.5.1)+(5.3.1) = 15+15 = 30 event space sample space drawing any 2 M&M's of the same color? sample space = = = 0.4545 0 12c event space =3c, 5c, c + 5c, 3c c [BY RD] G#d Ro] [B G= (R=D] =(3.1.1) + (10.1-1)+(11-6) event space sample space >>> 19 = 0.2879 66 = 3+10+6 =19 Drawing any 2 M&M's that are NOT all the same? (Hint: there's a hard way and an easy way to answer this question - both equally correct)