Question

at a convention of comedians, there were jokesters (j), tricksters (t), and pranksters (p). of those in attendance, 70 were jokesters. there were— 20 who were both t and j but not p. 20 who were t but neither j nor p. 35 who were p but neither t nor j. 15 who were both t and p but not j. 30 who were j but neither t nor p. 20 who were j and p but not t. how many comedians were at the convention?

Explanation:

Step1: Identify the non - overlapping groups

We have the following non - overlapping groups of comedians:

1. Those who are only $T$ and $J$: $n(T\cap J\cap\overline{P}) = 20$
2. Those who are only $T$: $n(T\cap\overline{J}\cap\overline{P})=20$
3. Those who are only $P$: $n(\overline{T}\cap\overline{J}\cap P)=35$
4. Those who are only $T$ and $P$: $n(T\cap\overline{J}\cap P)=15$
5. Those who are only $J$: $n(\overline{T}\cap J\cap\overline{P}) = 30$
6. Those who are only $J$ and $P$: $n(\overline{T}\cap J\cap P)=20$

Step2: Sum up the groups

The total number of comedians $N$ is the sum of all these non - overlapping groups.
$N=20 + 20+35 + 15+30+20$
$N=(20 + 20)+35+(15 + 30)+20$
$N = 40+35 + 45+20$
$N=(40+35)+(45 + 20)$
$N=75+65$
$N = 140$

Answer:

140