Question

A dinner party is to be fixed for a group of 100 persons. In this party, 50 people did not prefer fish, 60 preferred chicken, and 10 did not prefer either chicken or fish. How many people prefer both chicken and fish?
A. 20
B. 22
C. 25
D. None of these

Answer 1

Hint: The relationship between the number of elements in two sets, their union and intersection is given by \[n(A \cap B) = n(A) + n(B) - n(A \cap B)\]. First, we need to find the number of persons who prefer fish and the number of people who do prefer either fish or chicken.

Formula used
\[n(A \cap B) = n(A) + n(B) - n(A \cap B)\]

Complete step-by-step solution:
Given that the number of people are 100.
i.e., \[n({\rm{Total}}) = 100\]
Let the people who prefer fish be denoted by F
And the people who prefer chicken be denoted by C
Then, the number of people who do not prefer Fish is \[ = n(F') = 50\]
The number of people who prefer fish \[n(F)\]
\[ = n({\rm{Total}}) - n(F')\]
\[\begin{array}{l} = 100 - 50\\ = 50\end{array}\]
The number of people who prefer fish is 50.
Similarly, the number of people who do not prefer either fish or chicken is 10
\[ \Rightarrow \]The number of people who do prefer either fish or chicken is \[n(F \cap C) = 90\]
Then, we have the following data,
\[n(F) = 50,\;n(C) = 60\;{\rm{and}}\;n(F \cap C) = 90\]
\[n(F \cap C) = n(F) + n(C) - n(F \cap C)\]
Substituting the values
\[90 = 50 + 60 - n(F \cap C)\]
\[n(F \cap C) = 110 - 90\]
\[n(F \cap C) = 20\]
Hence, the number of people who prefer either Fish or Chicken are 20.
Therefore, the correct option is option A.

Note: Many students confused with the formula \[n(A \cap B) = n(A) + n(B) - n(A \cap B)\] and \[n\left( {A \cap B} \right) = n\left( A \right) + n\left( B \right)\]. When the events are independent then we use the formula \[n\left( {A \cap B} \right) = n\left( A \right) + n\left( B \right)\]. When the events are dependent then we use the formula \[n(A \cap B) = n(A) + n(B) - n(A \cap B)\].