. Total bulbs: n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs) = p(RR) Or p(BB) or p(GG)
$\frac{5}{16}$ * $\frac{4}{15}$ + $\frac{7}{16}$ * $\frac{6}{15}$ + $\frac{4}{16}$ * $\frac{3}{15}$
= $\frac{1}{12}$ + $\frac{7}{40}$ + $\frac{1}{20}$
= $\frac{10+21+6}{120}$ = $\frac{37}{120}$
(b) All different colors = 1-p (AIl the same colour) =
1 - $\frac{37}{120}$ = $\frac{83}{120}$
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
$\frac{5}{16}$ * $\frac{4}{15}$ + $\frac{5}{16}$ * $\frac{7}{15}$ + $\frac{5}{16}$ * $\frac{4}{15}$
= $\frac{1}{12}$ + $\frac{7}{48}$ + $\frac{1}{12}$
= $\frac{8+7}{48}$ = $\frac{15}{48}$
= $\frac{5}{16}$