If a fair sided die is rolled 3 times, what are the odds that neither

Question

If a fair sided die is rolled 3 times, what are the odds that neither a 2 or a 4 will show up on any roll?

1. 16/27

2. 8/27

3. 4/27

4. 2/27

5. 1/27

What does the term Odd mean here. If the question asks what is the probability that neither 2 or 4 will show up then I'm pretty sure the answer is 8/27.

But does odd mean -ve or oppose to ..
Please explain?

Thanks
-H

sag · Answer

Odds means Chances or Probability ..

IN such type of Q .. i take a contra approach..
As Q is asking neither a 2 or a 4 will show up on any roll
Therefore i take the probability of  either  2 or a 4 will show up on any roll = 2 /6
= 1 /3

Therefore Probability of neither a 2 or a 4 will show up on any roll = 1 - 1/3 = 2/3
As it is rolled 3 times : So probability = 2/3 * 2/3 * 2/3 = 8/27

neoreaves · Answer

Odds here is not the Even/Odds word ...this means probability

dixitraghav · Answer

the probablity of getting 2 or 4 is 4/6 = 2/3 in 1 chance so in 3 chance its (2/3)*(2/3)*(2/3) = 8/27 ans

ssgomz · Answer

help needed in understanding the concept

why is 1-((1/3)*(1/3)*(1/3)) wrong?

Bunuel · Answer

First of all odds and probability are not the same thing. If the probability of a certain event is  p , then the odds (odds in favor) of this event are  \frac{p}{1-p} . GMAT always asks about probability and not odds, so don't worry about this.

Now the probability that during 3 rolls neither 2 nor 4 will occur:  (\frac{2}{3})^3=\frac{8}{27} .

Odds in favor of this even would be  \frac{p}{1-p}=\frac{8}{19}  (8 to 19).

As  \frac{8}{19}  is not in answer choices, then I believe question means probability instead of odds.

Not a good question.

I guess you wanted to calculate the opposite probability and subtract it from 1. But  (\frac{1}{3})^3  is the probability that on all 3 rolls we'll get either 2 or 4, which is not the opposite of getting neither 2 nor 4 during 3 rolls.

The opposite probability of getting neither 2 nor 4 during 3 rolls is the sum of the following probabilities: getting 2 or 4  once  during 3 rolls,  twice  during 3 rolls, or  all three times  during 3 roll (that was the one you calculated).

Probability of getting 2 or 4  once  during 3 rolls is:  3*\frac{1}{3}*\frac{2}{3}*\frac{2}{3}=\frac{12}{27} , we are multiplying by 3 as we can get 2 or 4 on first, on second or on third roll;
Probability of getting 2 or 4  twice  during 3 rolls is:  3*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}=\frac{6}{27} , we are multiplying by 3 as this scenario also can occur in 3 ways: 2 or 4 on first and second, on first and third, or on second and third rolls;
Probability of getting 2 or 4  all three times  during 3 roll is:  \frac{1}{3}*\frac{1}{3}*\frac{1}{3}=\frac{1}{27} .

The opposite probability is the sum of above 3 probabilities:  P(opposite \ event)=\frac{12}{27}+\frac{6}{27}+\frac{1}{27}=\frac{19}{27} .

P(event)=1-P(opposite \ event)=1-\frac{19}{27}=\frac{8}{27}

Hope it's clear.

If a fair sided die is rolled 3 times, what are the odds that neither

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If a fair sided die is rolled 3 times, what are the odds that neither

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