Question

There are seven white and one brown eggs in an egg box. Ruchira decides to make a two-egg omelette. She takes out each egg from the box without looking at its colour. What is the probability that Ruchira gets an omelette made from 

(i) two white eggs 

(ii) one white and one brown egg 

(iii) two brown eggs

Answer 1

(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.