5B 4N 6R (15 bolas) a) $P(BB) = \frac515 \cdot \frac414= \frac20210 = \frac221$ b) $P(igual \: color) = P(BB) + P(NN) + P(RR) = $ $=\frac515 \cdot \frac414+\frac415 \cdot \frac314+\frac615 \cdot \frac514 = \frac20+12+30210=\frac62210=\frac31105$ c) $P(distinto \: color) = 1 - P(igual \: (…)