Let $t$ be TNYWR.

There are $t$ cards. Each card has one picture printed on it. The picture on each card is either of a dinosaur or a robot. Each dinosaur and robot is coloured either blue or green.

- 16 cards have blue dinosaurs
- 14 cards have green robots
- 36 cards have blue robots

If a card is chosen at random, what is the probability, written as a fraction in lowest terms, that it has either a green dinosaur or a blue robot printed on it?

## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca  2021 Canadian Team Mathematics Contest 
There are four possibilities for the picture printed on each card: a blue dinosaur, a blue robot, a green dinosaur, or a green robot.

Let $x$ denote the number of green dinosaurs.

From the information given, we have $14+16+36+x=t$ or $x=t-66$.

The number of cards with a blue robot printed on them is 36 , so the probability we seek is $\frac{x+36}{t}=\frac{t-66+36}{t}=\frac{t-30}{t}$.

Since $t=93$, the probability is $\frac{93-30}{93}=\frac{63}{93}=\frac{21}{31}$.

ANSWER: $\left(7,93, \frac{21}{31}\right)$