A survey of 500 television viewers produced the following information,285 watch football,195 watch hockey,115 watch basketball,45 watch football and
basketball,70 watch football and hockey,50 watch hockey and basketball and 50 do not watch any of the three games.
1)How many watch all the three games?
2)How many watch exactly one of the three games?
reply fast! tmr is my maths test.
To find the answer to these questions, we can use the principle of inclusion-exclusion.
1) To find the number of people who watch all three games, we need to subtract the number of people who do not watch any of the games from the total number of people who watch all three games.
Given information:
Total viewers (n) = 500
Not watching any game (n(A'UB'UC')) = 50
Therefore, the number of people who watch all three games (n(A∩B∩C)) can be calculated using the formula:
n(A∩B∩C) = n(A∪B∪C) - n(A'∪B'∪C')
We have the following information:
n(A∪B∪C) = 285 (watching football) + 195 (watching hockey) + 115 (watching basketball) = 595
Plugging in these values, we get:
n(A∩B∩C) = 595 - 50 = 545
Therefore, 545 people watch all three games.
2) To find the number of people who watch exactly one of the three games, we need to add up the number of people who watch only one game (football or hockey or basketball) while subtracting those who watch two games (football and hockey, football and basketball, or hockey and basketball).
The number of people who watch only football (n(A∩B'∩C')) can be calculated by subtracting the number of people who watch football and basketball (n(A∩B∩C')) and the number of people who watch football and hockey (n(A∩B∩C'). Similarly, we can find the number of people who watch only hockey (n(A'∩B∩C')) and only basketball (n(A'∩B'∩C)).
Using the principle of inclusion-exclusion and the given information, we can calculate the number of people who watch exactly one game:
n(A∩B'∩C') = n(A) - n(A∩B∩C') - n(A∩B∩C') = 285 - 45 - 50 = 190 (watching only football)
n(A'∩B∩C') = n(B) - n(A∩B∩C') - n(A∩B∩C') = 195 - 45 - 50 = 100 (watching only hockey)
n(A'∩B'∩C) = n(C) - n(A∩B∩C') - n(A∩B∩C') = 115 - 45 - 50 = 20 (watching only basketball)
Adding up these values, we get:
n(A∩B'∩C') + n(A'∩B∩C') + n(A'∩B'∩C) = 190 + 100 + 20 = 310
Therefore, 310 people watch exactly one of the three games.
I hope this explanation helps you understand how to solve these types of problems. Good luck with your math test!
u can try to explain logically and easily but I think this’ll not help me
I assume you did this with Venn diagrams
We don't know how many watch all three, label that x
look at the intersection of the hockey and Football circles, it says 70 watch both
but we already have x accounted for, so place 70-x in the part showing only hockey and football
insert 45-x and 50-x in the other overlaps in the same way.
Now look at the hockey circle
We already have entries of x, 70-x, and 50-x
so the part showing ONLY hockey = 195 - x - (70-x) - (50-x)
= 75+x
repeat for the other two circles
now add them all up and set equal to 500
solve for x (it better be a whole positive number)
now you can fill in all the values, and all mysteries will be revealed.
Just realized looking at the Related Questions below, that I answered this same question for you back in July of this year.
What gives ????