(a)Â Â Â P (all B+) = (0.12)(0.12)(0.12)=0.001728=0.1728%
(b)Â Â P (no B+) = (0.88)(0.88)(0.88) = 0.681472=68.15%
(c)Â Â Â P (not one B+) + P (no B+) = 1
P (not one B+) = 1 – P (no B+)
                           = 1 – 0.681472
P (at least 1 B+) = 0.318528=31.85%
(d)Â Â B. The event in part (a) is unusual because its probability is less than or equal to 0.05.