Question

10. The table below shows the probabilities of all realizations for each of the three dice. Notice that the dice are all "loaded” meaning that they are unbalanced in some way leading to an unequal probability of any number coming up. - 1 2 3 4 5 6 Die 1 6.0% 24.4% 11.4% 20.3% 20.7% 17.2% Die 2 17.5% 17.1% 14.9% 21.0% 16.5% 12.9% Die 3 7.9% 17.6% 20.9% 0.6% 21.2% 31.8%

Answer 1

ut x, y&z be the outcome of die 1, 2 and 3 respectively, Given Plaza)=6% - 0.06, P2 = 2): 0-244, Rope P(x=3)-0.114, PO4) = 0.0203&so on. P(minimum of x anda is 2) =P (,)- (2,2), (2,3)--- (2,6),(3,2),(4,2)--(6,2) - P(x=2, 272))+ P(272,8-2) +P(x=2,232) - P(x=2). PlzY2)+POW>2) P[Z=2)+P(21=2) Plz=2) - 0.244 (0.209 +0.006+0.212+0.318) + (0.114 +0.203+0.207+0,172)0.176+0.244*0-176 - 0.244 x 0.745 +0.696x0.176+0.244x0.176 = 0.18178 +0.122496+0.042944=0.34722 (minimum of x and t is 5) =P((3,2)=(5,6),(6,5),(5,5)) Pla:) P(2=6) +P(2-6) P12=5) + P(2=5) P12=5) = 0.207*0.318+0.17220.212+0.207x0.212 30.065826+ 0.036464+0.043884-10.146174 ① (minimum of x anda is less than 4) Pa 1 - P(both x and z is at least 4) -1- P(274).Plz24). = 1- (0.203+0.207+0.172) (0.006+0.212+0-318) -1-0.582x0.536 = (0.688048] Plmasumum of y and z is greater than 6) - [ol (i none of x, y, z can be more than6).