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I will challenge the next gym leader to a 3 vs 3 Pokemon battle.
I have 10 Pokemons all in all. Out of the ten, there are non-fire type Pokemons. The next gym leader has pure Grass types, and surely I may just need three fire type pokemons in my roster to defeat the gym leader.
But with the gym leader seeing that three fire-type Pokemons in my Pokemon roster is a disadvantage, he asked me to draw three pokeballs RANDOMLY instead, from my pool of 10 Pokemons, thus removing the certainty of facing three fire-type Pokemons.
The probability that none of the three Pokemons is a fire-type is 1/120.
I have a higher chance of defeating the Gym leader if I am able to draw even only one fire-type Pokemon. The probability that exactly one of my three randomly chosen Pokemons is a fire-type, is a fraction where and are relatively prime positive integers.
What is ?
(Image source: pokecommunity.com)
The answer is 47.
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1 solution
Let x be the number of fire-type Pokemon
For the probability that none of the three Pokemon is a fire-type, we need to choose three from the non-fire type, that is C(x,3). Then, the number of ways of choosing any three Pokemon from ten is C(10,3).
Thus, C(x,3)/C(10,3)=1/120 Solving x=3 -> non-fire type 10-x=7 -> fire type
The probability that exactly one of the three is a fire-type is (C(7,1)*C(3,2))/C(10,3)=7/40=a/b a+b=47 C(7,1) -> denotes choosing exactly one fire type Pokemon C(3,2) -> denotes choosing exactly two non-fire type Pokemon C(10C3) -> denotes choosing three Pokemon out of the ten