Question

A survey of $500$ television viewers produced the following information, $285$ watch football, $195$ watch hockey, $115$ watch basket-ball, $45$ watch football and basket ball, $70$ watch football and hockey, $50$ watch hockey and basket ball, $50$ do not watch any of the three games. The number of viewers, who watch exactly one of the three games is

• A. $325$
• B. $310$
• C. $405$
• D. $372$

Answer 1

The correct option is A $325$

Let U = set of television viewers, F = set of football viewers, H = set of hockey viewers and B = set of basket-ball viewers.

$\therefore n\left(F\right)=285,n\left(H\right)=195,n\left(B\right)=115,n(F\cap B)=45,$

$n(F\cap H)=70,n(H\cap B)=50,n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=50$

We have, $n(F^{\prime }\cap H^{\prime }\cap B^{\prime })=n\left(U\right)-n(F\cup H\cup B)$

$\therefore 50=500-n(F\cup H\cup B)\Rightarrow n(F\cup H\cup B)=500-50=450$

Also, $n(F\cup H\cup B)=n\left(F\right)+n\left(H\right)+n\left(B\right)-n(F\cap H)-n(H\cap B)-n(F\cap B)+(F\cap H\cap B)$

$\Rightarrow 450=285+195+115-70-50-45+n(F\cap H\cap B)$

$\Rightarrow 450=430+n(F\cap H\cap B)$

$\Rightarrow n(F\cap H\cap B)=450-430=20.$

(i) Number of viewers of all games $=n(F\cap H\cap B)=20$.

(ii) Number of viewers who watch exactly one game $=n\left(\right(F\cap H^{\prime }\cap B^{\prime })\cup (F^{\prime }\cap H\cap B^{\prime })\cup (F^{\prime }\cap H^{\prime }\cap B\left)\right)$

$=n(F\cap (H^{\prime }\cap B^{\prime }\left)\right)+n(H\cap (F^{\prime }\cap B^{\prime }\left)\right)+n(B\cap (F^{\prime }\cap H^{\prime }\left)\right)$

$=n(F\cap (H\cup B{)}^{\prime })+n(H\cap (F\cup B{)}^{\prime })+n(B\cap (F\cup H{)}^{\prime })$

$=\left[n\right(F)-n(F\cap (H\cup B)\left)\right]+\left[n\right(H\left)n\right(H\cap (F\cup B)\left)\right]+\left[n\right(B)-n(B\cap (F\cup H)\left)\right]$

$=\left[n\right(F)-n((F\cap H)u(F\cap B)\left)\right]+\left[n\right(H)-n((H\cap F)u(H\cap B)\left)\right]+\left[n\right(B)-n(B\cap F\left)u\right(B\cap H\left)\right)]$

$=n\left(F\right)-\left\{n\right(F\cap H)+n(F\cap B)-n((F\cap H)n(F\cap B)\}+n(H)-n(H\cap F)+n(H\cap B)-n\left(\right(H\cap F\left)n\right(H\cap B\left)\right)+n(B)-n(B\cap F)+n(B\cap H)-n\left(\right(BF\left)n\right(B\cap H\left)\right)=n(F)+n(H)+n(B)-2\{n(F\cap H)+n(F\cap B)+n(H\cap B)\}+3n(F\cap H\cap B)$

$=285+195+115-2(70+45+50)+3\left(20\right)=325.$