(i) P (first white egg) =\(\frac{7}{8}\) (7 white eggs, 8 total eggs)
P(second white egg) = \(\frac{6}{7}\) (6 white eggs, 7 total eggs as 1 egg has been taken out)
\(\therefore\) P ( omelette of two white eggs) = \(\frac{7}{8}\times\frac{6}{7}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)
(ii) P (first white egg) =\(\frac{7}{8}\)
P (second brown egg) =\(\frac{1}{7}\) |
P (first brown egg) = \(\frac{1}{8}\)
P (second white egg) = \(\frac{7}{7}\) = 1 |
\(\therefore\) P ( omelette of one white and one brown egg) = \(\frac{7}{8}\)x \(\frac{1}{7}\) + \(\frac{1}{8}\) x \(\frac{7}{7}\)
= \(\frac{1}{8}\) + \(\frac{1}{8}\) = \(\frac{1}{4}\)
Note. The eggs can be taken out in any order, so we consider both the cases, it is either first case or second case.
(iii) P ( first brown egg)= \(\frac{1}{8}\)
P (second brown egg)= \(\frac{0}{7}\)
\(\therefore\) P (omelette of two brown eggs) = \(\frac{1}{8}\) x 0 = 0.