Question

There are seven white and one brown egg 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. Choose the correct answer.
A. (i) \[\dfrac{3}{4}\] (ii) \[\dfrac{1}{4}\] (iii) 0
B. (i) \[\dfrac{3}{8}\] (ii) \[\dfrac{1}{4}\] (iii) 0
C. (i) \[\dfrac{3}{4}\] (ii) \[\dfrac{1}{2}\] (iii) 0
D. (i) \[\dfrac{1}{4}\] (ii) \[\dfrac{3}{4}\] (iii) 0

Answer 1

Hint: There total 8 eggs, so number of ways of selecting any two eggs from these 8 eggs will be \[{}^{8}{{C}_{2}}=\dfrac{8!}{2!(6!)}=28\]ways. Also, the number of ways of selecting two white eggs will be \[{}^{7}{{C}_{2}}=\dfrac{7!}{2!(5!)}=21\]ways. Similarly, number of ways of selecting one white and one brown egg is \[{}^{7}{{C}_{1}}\times {}^{1}{{C}_{1}}=\dfrac{7!}{1!(6!)}\times \dfrac{1!}{1!(0!)}=7\]ways. So, now the probability is given as \[P(x)=\dfrac{number\,of\,favourable\,\,ways}{total\,ways}\]

Complete step-by-step answer:
In the question, it is given that there are seven white and one brown egg 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. So, we have to find the probability that Ruchira gets an omelette made from
(i) Two white eggs
(ii) One white and one brown egg
(iii) Two brown eggs.
Now, here there are a total 8 eggs, out of which seven are white and one is brown egg. So, here the total number of ways of selecting two eggs from total 8 eggs is given by the combination concept. Since we are doing the selection, the number of ways of selecting two eggs from total 8 eggs will be \[{}^{8}{{C}_{2}}=\dfrac{8!}{2!(6!)}=28\]ways.
(i) Now, here we will use the combination concept and first find the ways of selecting 2 white eggs from 7 white eggs. So, it is given as \[{}^{7}{{C}_{2}}=\dfrac{7!}{2!(5!)}=21\]ways.
Now, we know that the probability is given as \[P(x)=\dfrac{number\,of\,favourable\,\,ways}{total\,ways}\]. Here the number of favourable ways is 21 and the total number of ways is 28. So the probability of making omelette with two white eggs will be:
\[\begin{align}
  & \Rightarrow P(x)=\dfrac{number\,of\,favourable\,\,ways}{total\,ways} \\
 & \Rightarrow P(x)=\dfrac{21}{28} \\
 & \Rightarrow P(x)=\dfrac{3}{4} \\
\end{align}\]
So, the required probability is case (i) is \[\dfrac{3}{4}\].
(ii) Now, the next part says the one is white egg and one is a brown egg. So, we will find ways of selecting 1 white egg from 7 white eggs and 1 brown egg from 1 brown egg. Now, selecting 1 white egg is \[{}^{7}{{C}_{1}}=\dfrac{7!}{1!(6!)}=7\]ways. Also, the number of ways of 1 brown egg from 1 brown egg is \[{}^{1}{{C}_{1}}=\dfrac{1!}{1!(0!)}=1\]way. So the total ways of selecting 1 white eggs from 7 white eggs and 1 brown egg from 1 brown egg is \[{}^{7}{{C}_{1}}\times {}^{1}{{C}_{1}}=\dfrac{7!}{1!(6!)}\times \dfrac{1!}{1!(0!)}=7\]ways.
Now, we know that the probability is given as \[P(x)=\dfrac{number\,of\,favourable\,\,ways}{total\,ways}\]. Here the number of favourable ways is 7 and the total number of ways is 28. So, the probability of making omelette with 1 white egg and 1 brown egg will be:
\[\begin{align}
  & \Rightarrow P(x)=\dfrac{number\,of\,favourable\,\,ways}{total\,ways} \\
 & \Rightarrow P(x)=\dfrac{7}{28} \\
 & \Rightarrow P(x)=\dfrac{1}{4} \\
\end{align}\]
So, the required probability is case (ii) is \[\dfrac{1}{4}\].
(iii) Finally, we have to find the probability that the omelette is made from two brown eggs. Now, since there is only one brown egg, there is no way that omelette can be made from two brown eggs. So the required probability is 0. So, the required probability is case (iii) is \[0\].

Hence, the correct answer is option A. (i) \[\dfrac{3}{4}\] (ii) \[\dfrac{1}{4}\] (iii) 0

Note: It can be noted that the combination \[{}^{n}{{C}_{r}}\]will not give the probability. It only gives the number of ways of selecting r things from the set of n things. Also, we need to simplify the fraction and write in the simplest form when finding the probability.