A survey of 500 television viewers produced the following information; 285 watch football, 195 watch hockey, 115 watch basketball, 50 do not watch any of the three games, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball.
(i) How many watch all the three games?
(ii) How many watch exactly one of the three games?
VerifiedHint:To solve this question, we will use the formula of three sets, that is, $n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-n\left( A\cap B \right)-n\left( B\cap C \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right)$. Also, we will use the formula for exactly one of the three games, that is, $n\left( A \right)+n\left( B \right)+n\left( C \right)-2\left[ n\left( A\cap B \right)+n\left( B\cap C \right)+n\left( A\cap C \right) \right]+3n\left( A\cap B\cap C \right)$. By using these formulas, we can find the solution of both the parts of the question.
Complete step-by-step answer:
In this question, we are asked to find the number of viewers who watch all the three games and who watch exactly one of the three games. We have been given that from a survey of 500 viewers, 50 do not watch any of the three games and therefore the total number of viewers are 500 – 50 = 450. Now, let us consider football viewers as the part of set A, hockey viewers as the part of set B and basketball viewers as the part of set C. So, from the given information in the question, that is, 285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball, we can say that,
$n\left(A \right )=285$.....(i)
$n\left(B \right )=195$ .....(ii)
$n\left(C \right )=115$ .....(iii)
$\left ( A\cap C \right )=45$.....(iv)
$\left ( A\cap B \right )=70$.....(v)
$\left ( B\cap C \right )=50$.....(vi)
Also, we have found out that the total viewers of game are 450. So, we can write it as $n\left( A\cup B\cup C \right)=450......\left( vii \right)$
(i) Number of viewers who watch all the three games, that is $n\left( A\cap B\cap C \right)$.
We know that for any three sets, we can apply the formula,
$n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-n\left( A\cap B \right)-n\left( B\cap C \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right)$
Now, we will put the values of each terms from the equations (i), (ii), (iii), (iv), (v), (vi) and (vii).So, we will get,
$450=285+195+115-70-50-45+n\left( A\cap B\cap C \right)$
Now, we will simplify the above expression, so we will get,
$n\left( A\cap B\cap C \right)=450-430=20......\left( viii \right)$
Hence, we can say that 20 viewers watch all three games.
(ii) Number of viewers who watch exactly one of the three games.
Here, we know that, for three sets,
n (exactly one game viewers) = $n\left( A \right)+n\left( B \right)+n\left( C \right)-2\left[ n\left( A\cap B \right)+n\left( B\cap C \right)+n\left( A\cap C \right) \right]+3n\left( A\cap B\cap C \right)$
Now, by substituting the values of the terms, we will get,
n (exactly one game viewers) = 285 + 195 + 115 – 2[70 + 50 +45] + 3(20)
n (exactly one game viewers) = 325
Hence, there are 325 viewers who watch exactly one of the three games.
Note: In this question, the most possible mistake one can make is in the calculation. So, the students are supposed to be focused while writing the formulas and doing the calculations. Also, the students can use the Venn diagram for finding the asked question by specially giving each part a number.
In the diagram, we have represented the number of viewers who watch only football as 1, only hockey as 2, only basketball as 3, football and hockey but not basketball as 4, hockey and basketball but not football as 5, football and basketball but not hockey as 6 and those who watch football, hockey and basketball as 7. By using these values, we can form equations based on the conditions given in the question and then solve the question. For example, if we consider the condition that 285 viewers watch football, then from the diagram we can form the equation that, n(1) + n(4) + n(7) + n(6) = 285. Similarly for all other conditions given, we can form equations and then solve them to get the answer.
