Rituraj has an aquarium, and he will buy exactly 2 of 5 creatures. The available creatures are: two turtles, a goldfish, a catfish and a sucker fish. It is given that a turtle does not fight with the other, neither does it attack the sucker fish. However the turtle will eat any fish in the aquarium. None of the other creatures fights with one another. What is the probability that if Rituraj randomly chooses two creatures, both of them will live umharmed?
A) 6/25
B) 3/5
C) 5/7
D) 1/2
E) Cannot be determined
Math
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The turtle does not .... attack the sucker fish. However the turtle will eat any fish in the aquarium.nayeem bari wrote:Rituraj has an aquarium, and he will buy exactly 2 of 5 creatures. The available creatures are: two turtles, a goldfish, a catfish and a sucker fish. It is given that a turtle does not fight with the other, neither does it attack the sucker fish. However the turtle will eat any fish in the aquarium. None of the other creatures fights with one another. What is the probability that if Rituraj randomly chooses two creatures, both of them will live umharmed?
A) 6/25
B) 3/5
C) 5/7
D) 1/2
E) Cannot be determined
Is that correct?
The turtle won't attack the sucker fish, but WILL eat it?
Rituraj's friend: Well it looks like the sucker fish was eaten last night.
Rituraj: Yes, but thankfully it wasn't attacked.
Cheers,
Brent
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Hi nayeem bari,
Brent brings up an interesting point (will a turtle harm the suckerfish or not). I'm going to assume that a turtle will be able to live with a suckerfish and NOT harm it.
The math behind this question can be done in a couple of ways. I'm going to use the Combination Formula and 'map' out the "harmful" options.
Since there are 5 creatures and we're going to choose 2 of them, the order does NOT matter (which means that we can use the Combination Formula to figure out the total number of options):
5c2 = 5!/(2!3!) = 10 different possible pairs of creatures
Now we have to figure out which pairs will lead to a 'harmful' result:
Turtle 1 and goldfish
Turtle 1 and catfish
Turtle 2 and goldfish
Turtle 2 and catfish
So 4 of the 10 options will lead to a 'harmful' result; this means that 6 out of 10 will have an 'UNharmful' result.
6/10 = 3/5
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
Brent brings up an interesting point (will a turtle harm the suckerfish or not). I'm going to assume that a turtle will be able to live with a suckerfish and NOT harm it.
The math behind this question can be done in a couple of ways. I'm going to use the Combination Formula and 'map' out the "harmful" options.
Since there are 5 creatures and we're going to choose 2 of them, the order does NOT matter (which means that we can use the Combination Formula to figure out the total number of options):
5c2 = 5!/(2!3!) = 10 different possible pairs of creatures
Now we have to figure out which pairs will lead to a 'harmful' result:
Turtle 1 and goldfish
Turtle 1 and catfish
Turtle 2 and goldfish
Turtle 2 and catfish
So 4 of the 10 options will lead to a 'harmful' result; this means that 6 out of 10 will have an 'UNharmful' result.
6/10 = 3/5
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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Hi nayeem bari,
Assuming the turtle does not attack the suckerfish, we have 3 possible outcomes
Outcome#1
Selecting 2 Turtles
i.e 2c2/5c2 = 1/10.
Outcome#2
Selecting 1 Turtle and 1 Suckerfish
2c1 * 1c1 / 5c2 = 1/10 (Remember : 2c1 is for selecting a turtle and 1c1 for suckerfish)
Outcome#3
Selecting 2 creatures other than turtles
3c2/5c2 = 3/10
So final probability = 1/10 + 1/5 + 3/10 = 3/5
Hope it helps.
Assuming the turtle does not attack the suckerfish, we have 3 possible outcomes
Outcome#1
Selecting 2 Turtles
i.e 2c2/5c2 = 1/10.
Outcome#2
Selecting 1 Turtle and 1 Suckerfish
2c1 * 1c1 / 5c2 = 1/10 (Remember : 2c1 is for selecting a turtle and 1c1 for suckerfish)
Outcome#3
Selecting 2 creatures other than turtles
3c2/5c2 = 3/10
So final probability = 1/10 + 1/5 + 3/10 = 3/5
Hope it helps.
Hi Guys
Why do you guys treat turtles as two different possible selection? Isn't it like dice problems?
Here is my approach:
total number of outcomes: 2c2 (for turtles)+3c1(turtle and other)+3c2(others) = 7
total safe outcomes: 2c2 (for turtles)+1c1 (super + turtle) + 3c2(others) =5
so the answer will be: 5/7
Why do you guys treat turtles as two different possible selection? Isn't it like dice problems?
Here is my approach:
total number of outcomes: 2c2 (for turtles)+3c1(turtle and other)+3c2(others) = 7
total safe outcomes: 2c2 (for turtles)+1c1 (super + turtle) + 3c2(others) =5
so the answer will be: 5/7
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As Rich demonstrates, we don't really need to get into a lot of probability/counting concepts here.Rituraj has an aquarium, and he will buy exactly 2 of 5 creatures. The available creatures are: two turtles, a goldfish, a catfish and a sucker fish. It is given that a turtle does not fight with the other, neither does it attack the sucker fish. However the turtle will eat catfish and goldfish. None of the other creatures fights with one another. What is the probability that if Rituraj randomly chooses two creatures, both of them will live umharmed?
A) 6/25
B) 3/5
C) 5/7
D) 1/2
E) Cannot be determined
Once we recognize that the TOTAL number of possible outcomes = 5C2 = 10, we know that there are, AT MOST, 10 scenarios in which no fish die (or do die, as in Rich's solution). This gives us the freedom to simply list those scenarios.
Here are all of the scenarios in which no fish die:
- Turtle 1 and turtle 2
- Turtle 1 and sucker fish
- Turtle 2 and sucker fish
- Goldfish and sucker fish
- Catfish and sucker fish
- Goldfish and catfish
There are 6 such outcomes
So, P(no fish die) = 6/10 = [spoiler]3/5 = B[/spoiler]
Cheers,
Brent
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Another (more probabilistic) way of solving this:
Rituraj's creatures will eat each other if he picks (i) a turtle and (ii) a catfish or a goldfish.
There are two ways in which this could happen. (This next step is very tricky, but is important in probability. When a scenario could happen in more than one order, consider both orders!)
1: Turtle first, unlucky fish second.
2: Unlucky fish first, turtle second.
P(1) = (2/5) * (2/4) = 2/10
P(2) = (2/5) * (2/4) = 2/10
Adding this up, we get 4/10, or 2/5. So there's a 2/5 chance the turtle eats one of the fish, meaning there is a (1 - 2/5), or 3/5 chance that harmony reigns in the aquarium. (Let's hope!)
Rituraj's creatures will eat each other if he picks (i) a turtle and (ii) a catfish or a goldfish.
There are two ways in which this could happen. (This next step is very tricky, but is important in probability. When a scenario could happen in more than one order, consider both orders!)
1: Turtle first, unlucky fish second.
2: Unlucky fish first, turtle second.
P(1) = (2/5) * (2/4) = 2/10
P(2) = (2/5) * (2/4) = 2/10
Adding this up, we get 4/10, or 2/5. So there's a 2/5 chance the turtle eats one of the fish, meaning there is a (1 - 2/5), or 3/5 chance that harmony reigns in the aquarium. (Let's hope!)