Question

A survey of '500 television watchers produced the following information; 285 watch foot-ball, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball , 50 do not watch any of the three games. How many watch all the three games ? How many watch exactly one of the games?

Answer 1

Only football 190 etc.Given N=500, n (F)=285,$n\left(H\right)=195,n\left(B\right)=115,n(F\cap B)=45,$ $n(F\cap H)=70,n(H\cap B)=50,$ $n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=50.$ To find $n(F^{\prime }\cap H\cap B),n(F\cap H^{\prime }\cap B^{\prime }),$ $n(F^{\prime }\cap H\cap B^{\prime }),n(F^{\prime }\cap H^{\prime }\cap B)$ we have $50=n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=n(F\cup H\cup B){,}^{\prime }$ By De-Morgan law $=N-n(F\cup H\cup B)=N-\{S_1-S_2+S_3\}$ $=N-\{n\left(F\right)+n\left(H\right)+n\left(B\right)-n(F\cap H)-n(H\cap B)-n(B\cap F)+n(F\cap H\cap B)\}$ $=500-285-195-115+70+50+45-n(F\cap H\cap B)$ $=665-595-n(F\cap H\cap B)$ $=70-n(F\cap H\cap B)$ $\therefore n(F\cap H\cap B)=20$ Again, $n(F\cap H^{\prime }\cap B^{\prime })=n\{F\cap {(H\cup B)}^{\prime }\}$ by De-Morgan law $=n\left(F\right)-n\{F\cap (H\cup B)\}$ $=n\left(F\right)-n\{(F\cap H)\cup (F\cap B)\}$ $=n\left(F\right)-\{n(F\cap H)\cup (F\cap B)\}$ $=n\left(F\right)-\{(F\cap H)+n(F\cap B)-n(F\cap H\cap B)\}$ $=285-70-45+20=190$