An oddly shaped die has N sides, numbered from 1 to N, that are equall

P = 1 - P (no "1" in two throws)

P (no "1" in two throws) = \((\frac{N-1}{N})^2\)
=> \(P = 1 - (\frac{N-1}{N})^2\)

\(P = 1 - (\frac{N-1}{N})) * (1+(\frac{N-1}{N})) = \frac{2N-1}{N^2}\)

(1) N is prime
N=2, \(P = \frac{3}{4} (> \frac{1}{5})\)

N=7, \(P = \frac{13}{49} (> \frac{1}{5})\)

N=11, \(P = \frac{21}{121} (< \frac{1}{5})\)

Thus, (1) alone is INSUFFICIENT

(2) N>5
For N=7 & N = 11, we have two different possibilities, thus, (2) alone is INSUFFICIENT

(1) & (2) together are INSUFFICIENT as well since 7 & 11 are prime and >5 and yet present different possibilities.

Thus, Answer is E.

This method however, took me >3 min to solve the problem. Could anyone please suggest a shorter method?

Hi chetan2u,

You ask a good question; one that's rooted more in 'grammar' than 'math.'

Here, the modifier "oddly" describes the "shape" of the die, not the number of sides. This is the writer's way of saying "yes, it's weird that the die might have 2 or 3 sides, for example, instead of the standard 6 sides..."

If this were an Official GMAT question AND the description was meant to be about the NUMBER of sides, then the wording would likely have been "a die with an odd number of sides...." (or equivalent).

While wording on certain practice questions from GMAT Prep Companies will vary in quality and similarity to the Official material, the questions that are written by GMAC writers are always carefully crafted. One of the goals of those same writers is to remove all possibility of "bias" on the part of the reader.

GMAT assassins aren't born, they're made,
Rich

MAGOOSH OFFICIAL SOLUTION:

First of all, if the N sides are equally likely, the probability of getting “1” on a single throw, or any particular side on a single throw, is 1/N. The probability of not getting “1” on a single throw, or not getting any particular side, is (1 – 1/N) = (N – 1)/N.

For relatively low values of N, then getting “1” is relatively more likely, and getting something else is relatively less likely. As N gets bigger, the probability of getting “1” on a single throw decreases, and the probability of getting something else increase.

What really makes the difference in this probability is the size of N. This is an at least probability, so we calculate the probability of getting something other than one on two consecutive throws, and then subtract this from 1.

Statement #1: this statement is not very helpful. If N = 2, then essentially, the “two-sided die” is a coin, and if H = 1, then the probability of not getting a “1” is (1/2). The probability of not getting a “1” on two consecutive throws is (1/4). That’s the probability of getting no “1” on two throws. We subtract this from one to get the probability of at least one “1”: P = (3/4), which greater than 1/5. Answer to the prompt = “no.”

But, suppose N is larger. For easy of calculation, I will pick N = 15, even though 15 is not prime. If N = 15 produces a probability less than 1/5, then so will any prime number greater than 15.

If N = 15, then the probability of not getting a “1” on a single throw is 14/15. The probability of not getting a “1” on two consecutive throws is this squared, 196/225. If this is more than 4/5, then subtracting it from one will be less than 1/5. Well, 225/5 = 45, and 45 times 4 is 180. Thus, (180/225) = (4/5), and (196/225) > (4/5). Subtracting this from one will produce a probability less than 1/5. Answer to the prompt = “yes.”

This statement is consistent with different answers to the prompt question, so it does not lead to a definitive answer. This statement, alone and by itself, is not sufficient.

Statement #2: We have already seen that N = 15, or any number larger than that, produces a “yes” answer to the prompt. Is it possible to produce a “no” answer to with a value of N greater than 5?

I will use N = 7. The probability of not getting a “1” on a single throw is 6/7. The probability of not getting a “1” on two consecutive throws is this squared, 36/49. If this is less than 4/5, then subtracting it from one will be more than 1/5.

Well, clearly, 36/45 = 4/5, and making the denominator larger makes a fraction smaller, so

4/5 = 36/45 > 36/49. So, this fraction is less than 4/5, so subtracting it from one will result in a probability that is greater than 1/5. This gives a “no” answer to the prompt.

This statement is consistent with different answers to the prompt question, so it does not lead to a definitive answer. This statement, alone and by itself, is not sufficient.

Combined statements: As we saw in the previous statements, the prime number N = 7 gives a “no” answer, and the non-prime N = 15 gives a “yes” answer, which implies that any prime number greater than 15 (such as 17, 19, 23, etc.) would also give a “yes” answer. Different answers are possible even with the constraints of both statements. Together, both statements are still not sufficient.

Answer = (E)

Hello @Bunnel, Thanks for the explanation however when i solved it i solved it in below way, which gave me definite answer.

Could you please correct me, where i am getting wrong.

Problem statement :
An oddly shaped die has N sides, numbered from 1 to N, that are equally likely to appear on a throw. Let P be the probability that at least one “1”
Let P(X) be the probability of getting 1,

My Attempt
Let P(X’) be the probability

P(X) = 1/N
P(X’) = 1 – (1/N) = (N-1)/N
Let P(Y) = probability of getting at least one “1”
P(Y) = P(One “1” and another no “1”) + P(Both 1)
P(y) = P(X)*P(X’) + P(X)*P(X) = {(1/N)*(N-1)/N) } + {(I/N) * (1/N)}
P(Y) = 1/N

Now question asks if P(Y) < 1/5
Can be further simplified to
1/N < 1/5
5 < N

So statement II, answers whether N is greater than 5 or not.
Please guide, where I am getting wrong.

Dear vishwaprakash

The mistake you made was in the equation for P(Y).

The correct equation will be: P(Y) = P(1 on 1st throw and another number on 2nd throw) + P(another number on 1st throw and 1 on 2nd throw) + P(1 on both throws)

So, P(Y) = \(\frac{1}{N}*\frac{(N-1)}{N}\) + \(\frac{1}{N}*\frac{(N-1)}{N}\) + \(\frac{1}{N} * \frac{(1}{N)}\) = \(\frac{(2N-1)}{N^2}\)

Hope this helped!

Best Regards

Japinder

Hello @Japinder

Thanks for the explanation, I got my mistake. I forgot to consider one of the possibility.

Regards,
Vishwaprakash

P(zero "1" appears in one throw) = \(\frac{N-1}{N}\)
P(zero "1" appears in two throws) = \(\left(\frac{N-1}{N}\right)^2\)

The original question: Is \(1-\left(\frac{N-1}{N}\right)^2<\frac{1}{5}\) ?

We can solve the above inequality and rephrase the question accordingly.

\(\frac{N^2-N^2+2N-1}{N^2}<\frac{1}{5}\)

\(10N-5<N^2\)

\(N^2-10N+5>0\)

\(N^2-10N+25-25+5>0\)

\((N-5)^2>20\)

\(|N-5|>2\sqrt{5}\approx 2\cdot 2.25=4.5 \implies\) N<0.5 OR N>9.5

Since N must be an integer greater than or equal to 1, the relevant part of the rephrased question is simple:
Is N>9.5 ?

1) We know that N is a prime. If N=7, then the answer to the rephrased question is No. However, if N=11, then the answer to the rephrased question is Yes. Thus, we can't get a definite answer to the rephrased question. \(\implies\) Insufficient

2) We know that N>5 and could test the same cases that we used to test 1). \(\implies\) Insufficient

1&2) Since we can use the same cases to prove that statement 1) and 2) are insufficient, 1&2) must also be insufficient. \(\implies\) Insufficient

Answer: E

Using both statements, N can be huge. Then P will be minuscule, so P can be much less than 1/5. We'll get the largest value of P when N is as small as possible, i.e. when N = 7. Then the probability we don't get a '1' is (6/7)(6/7) = 36/49, and the probability we do is 13/49, which is greater than 13/50 = 0.26. So we can get yes and no answers to the question, and the answer is E.

I share chetan's objection to the phrase "oddly shaped", which someone might reasonably think is a phrase with a special but unfamiliar mathematical definition.

oddly shaped= 3,5,7,9, etc.

Probability of 1 appears = [1- ( 1-n)/n*(1-n/n)]= 2n-1/n^2

1.) N = 2,3,5,7,11 etc.
not sufficient

2.) N>5
n can be 6,7,8,9 etc.

say n = 6; probability = 11/36 compare with 1/5( 7.xx/36)- P>lower
n = 7; probability = 13/49 compare with 9.8/49 p>(1/5)
n= 11; probability 21/121 compared with 1/5( 24.2/121) p<1/5
n = 100 ; probability = 199/10000 compare with 1/5( 2000/10000) p< 1/5

it is not always higher than 1/5
not Sufficient

1+2: 7,11
not sufficient
hence E

Can someone explain why the probability for n = 7 is p = 13/49?

I got 1/7 x 6/7 = 6/49 <--- Probability of getting one on the first roll is 1/7 and "not one" on the second roll is 6/7

When I look at this probability it makes me think about the probability of NOT getting 1, rather than getting 1.

P(zero "1" appears in one throw) = N−1 / N
P(zero "1" appears in two throws) = (N−1/N)^2

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