Question

\[125\] aliens descended on a set of film as extra-terrestrial beings. \[40\] had two noses. \[30\] had three legs. \[20\] had four ears. \[10\] had two noses and three legs. \[12\] had three legs and four ears. \[5\] had two noses and four ears. \[3\] had all the three features. How many were without any usual feature.
A. \[3\]
B. \[35\]
C. \[80\]
D. None of these

Answer 1

Hint: Here we take each feature as an event and draw a Venn diagram to find the solution. We find the union of all three events as we are given all the intersections of events and subtract that value of union of events from the total number of aliens.
* We know if \[A,B,C\] are three events then we can write
\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\]

Complete step-by-step answer:
Total number of aliens descended on a set of films are \[125\].
We are given the data
\[40\] had two noses.
\[30\] had three legs.
\[20\] had four ears.
\[10\] had two noses and three legs.
\[12\] had three legs and four ears.
\[5\] had two noses and four ears.
\[3\] had all the three features.
Let us denote the feature of two noses by \[X\], denote the feature of three legs by \[Y\] and the feature of four ears by \[Z\].
Then we can draw a Venn diagram depicting the situation.

We can write
\[40\]had two noses. So, \[P(X) = 40\]
\[30\] had three legs. So, \[P(Y) = 30\]
\[20\] had four ears. So, \[P(Z) = 20\]
\[10\] had two noses and three legs. So, \[P(X \cap Y) = 10\]
\[12\] had three legs and four ears. So, \[P(Y \cap Z) = 12\]
\[5\] had two noses and four ears. So, \[P(X \cap Z) = 5\]
\[3\] had all the three features. So, \[P(X \cap Y \cap Z) = 3\]
Therefore we can find the value of \[P(A \cup B \cup C)\] by substituting the values in the equation.
\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\]
\[
  P(A \cup B \cup C) = 40 + 30 + 20 - 10 - 12 - 5 + 3 \\
  P(A \cup B \cup C) = 93 - 27 \\
  P(A \cup B \cup C) = 66 \\
 \]
Now we find the number of aliens without any of the three features by subtracting the union value from the total number of aliens.
Therefore, number of aliens without any feature is \[125 - 66 = 59\]

Therefore, option D is correct.

Note: Students are likely to get confused with the number of features as it is given two noses, three legs and four ears but these numbers are not to be counted separately. Keep in mind while drawing the Venn diagram the three sets should intersect each other else we will not be able to show intersection sets.
Students many times make mistake of assuming union of all sets as the universal set, i.e. here they might assume that union gives the number of total aliens, but keep in mind that union set contains aliens which are in set X or in set Y or in set Z, but universal set contain all the aliens that landed on the set.