12 Days of Christmas GMAT Competition 2025 - Day 9: John is applying

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Bunuel

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25 Dec 2025, 10:00

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GMAT Club Official Explanation:

John is applying to be the leader of a study group. To qualify, he can either take a short test in which he must solve both of the 2 questions, or he can take a longer test in which he must solve at least 3 of the 4 questions. John’s probability of solving any one question correctly is p, where 0 < p < 1. Would John have a better chance of qualifying if he chose the longer test?

Short test qualifying probability is p^2.

Longer test qualifying probability is P(exactly 3 out of 4 correct) + P(all 4 correct) = 4!/3! * p^3(1 - p) + p^4 = 4p^3 - 3p^4.

The longer test is better when 4p^3 - 3p^4 > p^2:
4p - 3p^2 > 1
3p^2 - 4p + 1 < 0
(3p - 1)(p - 1) < 0
1/3<p<1

Therefore, the answer to the question is YES when p is greater than 1/3, and no when p is less than or equal to 1/3.

(1) The probability that John qualifies if he chooses the short test is less than 1/4.

This means p^2 < 1/4, so p < 1/2. p could still be below 1/3 or above 1/3. Not sufficient.

(2) The probability that John does not qualify if he chooses the short test is greater than 8/9.

Not qualifying on the short test is 1 - p^2. So, we get 1 - p^2 > 8/9, which gives p^2 < 1/9, thus p < 1/3. If p < 1/3, the longer test is not better. Sufficient.

Answer: B.

General Discussion

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25 Dec 2025, 09:49

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Short Test
P(ST Success) = p * p = p^2

Long test
P(LT Success) = p^4 + 4C3 * p^3(1-p) = p^4 + 4p^3 - 4p^4 =4p^3 - 3p^4

Q: Is P(LT) > P(ST)?

4p^3 - 3p^4 > p^2
4p -3p^2>1
3p^2 - 4p + 1 < 0
3p^2 -3p -p + 1 < 0
3p(p - 1) -1(p-1) <0
(3p - 1) (p - 1) < 0

Only 1/3 8/9
p^2 < 1/9
p < 1/3
Sufficient

Answer B

Bunuel

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25 Dec 2025, 10:03

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John is applying to be the leader of a study group.
To qualify, he can either take a short test in which he must solve both of the 2 questions, or he can take a longer test in which he must solve at least 3 of the 4 questions.
John's probability of solving any one question correctly is p, where 0<p<1.

Would John have a better chance of qualifying if he chose the longer test ?
The probability of passing the shorter test = p*p
The probability of passing the longer test = 4C1(1-p)*p*p*p + p^4 = 4p^3(1-p) + p^4 = p^3 (4 - 4p + p) = p^3(4-3p)

John would have better chance of qualifying if he chose the longer test if p^3(4-3p) > p^2
p^2 {p(4-3p) > 1}
4p - 3p^2 - 1 > 0
3p^2 - 4p + 1 <0
3p^2 - 3p - p + 1 < 0
3p(p-1) - (p-1) < 0
(3p - 1) (p -1 ) < 0
1/3 1/3: Long test is better
p< 1/3; Short test is better
p = 1/3; Both are same

(1) The probability that John qualifies if he chooses the short test is less than 1/4.
The probability that John qualifies if he chooses the short test = p^2 < 1/4
p < 1/2
Case 1: p = 1/4; He is better if he opts for shorter test
Case 2: p = 5/12; He is better if he opts for longer test
Not sufficient

(2) The probability that John does not qualify if he chooses the short test is greater than 8/9.
The probability that John does not qualify if he chooses the short test = 1 - p^2 > 8/9
p^2 < 1/9
p < 1/3
He is better if he opts for short test.
Sufficient

IMO B

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25 Dec 2025, 10:03

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In short he must solve both 2
(Provability of short test) X= p*p= p^2
In long test he must solve atleast 3 of 4
(Probability of solving long test) Y= (4/3)(p^3)(1-p)+(4/1)(p^4) = p^3 (4-3p)
Y>X ; (p^3)(4-3p)>p^2 ; (3p^2)-4p+1<0 ; (3p-1)(p-1)<0
(1/3)<p<1

(1) X<1/4 ; p^2 <1/4 ; p<1/2
p can be either beliw or above 1/3
Insufficient
(2) Not qualify short test>8/9
(1-p^2)>8/9 ; p^2 <1/9 ; p<1/3
Sufficient

B

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25 Dec 2025, 10:17

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25 Dec 2025, 10:20

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Bunuel

Let's see under which condition John will have a better chance of qualifying if he choses the longer test

p*p < 4C3 p^3(1-p)+p^4

Dividing by p^2 we get

1 < 4p(1-p)+p^2

1 < 4p - 3p^2

3p^2 - 4p + 1 < 0

(p-1)(3p-1) < 0

1/3 < p < 1

0.33 < p < 1

Hence, when p is between 0.33 and 1, John has better chances of acing the longer test.

1) p*p < 1/4

p < 1/2

p < 0.5

However we don't know if p < 0.33. Hence, this statement alone is not sufficient.

Eliminate A, and D.

2) Probability of not qualifying > 8/9

Therefore probability of qualifying is less than 1/9

p^2 < 1/9

p< 1/3

p < 0.33

Hence, this statement is sufficient to answer the question.

Option B

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25 Dec 2025, 10:39

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Bunuel

p(a) = 2 ques = p^2
p(b) = atleast 3 of 4 ques = p^4 + 4(1-p) (p^3) = 4p^3 - 3p^4

p(a) > p(b) = p^2 > 4p^3 - 3p^4
on solving we get

it for p<1/3

using s1) we get p <1/2
for which we can;t hget decisive set

using s2) we get p^2 < 1/9
p<1/3

hence this is enough
answer is B

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25 Dec 2025, 11:23

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first I found for what values of p the probability of passing with the long test were higher than with the short test.
When PLong > PShort?
Plong = probability of 3 right questions + probability of 4 right questions = 4C3 * p^3 * (1-p) + p^4 = 4p^3 -3p^4
Pshort = p^2

4p^3 - 3p^4 > p^2
3p^2 - 4p + 1 < 0
the equation is < 0 when 1/3 considering p must be <1 -> PLong>PShort when p>1/3

(1) not sufficient
if p^2<1/4 -> p<1/2
this could mean that p can be both > or < than 1/3

(2) sufficient
probability of passing = 1 - probability of not passing = 1 - 8/9 = 1/9
if p^2<1/9 -> p<1/3
this means the short test is better

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25 Dec 2025, 11:27

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Short test:
both correct P =p*p
Long test: either all correct or 3 correct, 1 wrong
probability = $p^4 + 4* p^3 * (1-p)$ = $4p^3 - 3p^4$

is long test better than short test??

$4p^3 - 3p^4$ > $p^2$
solving gives, iff 1/3<p<1..
so, is P>0.33??

1. $p^2 < 0.25$..p<0.5...so, p can be 0.2 or 0.4...NOT SUFFICIENT
2. p(1-p)+(1-p)p+(1-p)(1-p) > 8/9
p<0.33....SUFFICIENT

Ans B

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25 Dec 2025, 11:34

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Let's break the problem:

The probability of a single success is always p. Would he have a better probability of qualifying if he chooses the long test?

P( qualifying | short test) = p^2
P( qualifying | long test) = P( success at least 3 problems) = P( success at exactly 3 problems) + P( success 4 problems) = p^3*(1-p)*4!/3! + p^4 = 4p^3 - 4p^4 = 4p^2(p - p^2)

We have that P (qualifying | long test) > P (qualifying | short test) if 4p^2(p-p^2) > p^2
hence if p - p^2 > 1/4

I) P ( qualifying | short test) < 1/4 ==> p^2 < 1/4 ==> p < 1/2 NOT SUFFICIENT

II) P( qualyfying | short test) < 1/9 ==> p^2 < 1/9 ==> p< 1/3 and in this case it's SUFFICIENT

IMO B!

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25 Dec 2025, 11:36

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This question is one of the best !

p(qualifying | John takes short test) $= p * p = p^{2}$

p(not qualifying | John takes short test) $= 1 - p^{2}$

p(qualifying | John takes long test) = 3 pass or all 4 pass
$= 4c3 * p^{3} * (1-p) + p^{4}$

$= 4 * p^{3} - 4*p^{4} + p^{4}$

$= 4*p^{3} - 3*p^{4}$

question:
is p(qualifying | John takes short test) < p(qualifying | John takes long test) ?

$p^{2} < 4 * p^{3} - 3*p^{4}$

$3p^{4}-4p^{3}+p^{2} < 0$

$p^{2}(3p^{2}-4p+1) < 0$

$p^{2}(3p-1)(p-1) < 0 $

$\frac{1}{3} \frac{8}{9}$

$p^{2} < \frac{1}{9}$

$|p| < \frac{1}{3}$

since p is always between o and 1

$p < \frac{1}{3}$

sufficient

Ans: Option B

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25 Dec 2025, 11:44

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Bunuel

Let Ps = probability to qualify on short test
Let Pt = probability to qualify on long test
short test: must get 2/2, so Ps = p^2
Long test: must get atleast 3/4, so Pt = 4C3.p^3.(1-p) + 4C4.p^4 = 4p^3(1-p) + p^4 = 4p^3 - 3p^4

compare:
Pt > Ps => 4p^3 - 3p^4 > p^2 => p^2(4p - 3p^2 - 1) > 0
since 0<p<1, this reduces to 4p - 3p^2 - 1 > 0
3p^2 - 4p + 1 < 0 => (3p - 1)(p - 1) < 0
so Pt > Ps exactly when p > 1/3 ( and p < 1 already)

now test statements:
(1) Ps < 1/4 => p^2 < 1/4 => p < 1/2
this does not tell whether p > 1/3
not sufficient

(2) 1 - Ps > 8/9 => Ps < 1/9
p^2 < 1/9 => p < 1/3
then p > 1/3 is false, so longer test is not better
Sufficient

OPTION B

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25 Dec 2025, 11:54

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Bunuel

If $\frac{1}{x}$ represents the chance of John getting a question right, we essentially are trying to find a value for $\frac{1}{x}$, and seeing if it's higher if you have to get 2/2 questions right or 3/4 questions right. This is a yes or no question.

Statement 1 - put into an equation, this is:
$\frac{1}{x^2}<\frac{1}{4}$, which becomes $\frac{1}{x}<\frac{1}{2}$.

The range of $\frac{1}{x}$ being less than $\frac{1}{2}$ is vast though. We could have a probability very close to $\frac{1}{2}$ which would result in a value close to $\frac{4}{16}=\frac{1}{4}$ for the long test and $\frac{1}{4}$ for the short test; one is not better than the other. Or, we could have a value like $\frac{1}{10}$, where the probability of getting the long test right would be $\frac{9}{250}$, at which point the short test is better. Statement 1 therefore does not answer the question of "is the long test better?", and is insufficient.

Statement 2 - put into an equation, this becomes:
$1-\frac{1}{x^2}>\frac{8}{9}$. Simplified, this is $\frac{1}{x}<\frac{1}{3}$.

Let's test the value of 1/3, as we know from statement 1 that very small values like 1/10 prefer the short test vs long test.

$\frac{1}{3^2}=1/9$
$\frac{1}{3^3}*\frac{2}{3}*4=\frac{8}{81}$
$\frac{8}{81}<\frac{1}{9}$

We see again that the short test is better, even at 1/3. Since we already tested 1/10, all values smaller than 1/3 will prefer the short test. From Statement 2, the answer for the question "Is the short test better?" is a definitive yes. Our answer is B, Statement 2 is sufficient.

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25 Dec 2025, 14:34

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Probability of qualifying for the short test is pxp= p^2
Probability of qualifying for the long test is 4!/3!p^3(1-P)+p^4= 4p^3-3p^4
To test if the longer test offers a better chance?
4p^3-3p^4>p^2 which simplfies to form (3p-1)(p-1)<0. This case is only possible when p>1/3
S1 Translates into p<1/2 which is insufficient because many values exist which are both greater than and less than 1/3 hence insufficient
S2 This means 1-p^2>8/9 or p^2<1/9 and that p<1/3 meaning the long test does not offer a better chance hence sufficient
Ans B

Bunuel

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25 Dec 2025, 22:49

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each statement alone will be sufficient to find a unique test hence option d is correct

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25 Dec 2025, 23:23

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I have never seen such questions which are based on mixture of inequality and probability. First of all , hats off whoever made this question. Let's solve it:

Probability of solving a question correctly is p.
To get success in shorter test , one needs to solve both questions correctly.
So , P(shortest test success) = P(CC) = P(C)*P(C) = p^2

And , P(Not shorter test success) = 1-p^2

Similarly , to get success in longer test , one needs to solve at least 3 questions correctly.
P(longer test success) = P(CCCW)+P(CCCC) = P(C)^3*P(W)*4 + P(C)^4 [ Applying Probability of complex events concept]
=4 p^3*(1-p) + p^4 = 4p^3-3p^4

Also since statements are based on inequalities, let's also compare P(success in longer test) and P(success in shortest test)

P(success in longer test) - P(success in shorter test)
= 4p^3-3p^4-p^2
= p^2(4p-3p^2-1)
= -p^2(3p^2-4p+1)
=-p^2(p-1/3)(p-1)

So let's analyze the above equation , applying wavy curvy concept,

above one will be > 0 for p = (1/3,1)

for other values of p except p=1/3 ,1 it will be < 0.

Now let's check statement 1.

it says P(success in shorter test) <1/4
p^2<1/4
p<1/2

But as wee see that as long as p is in (1/3 ,1/2) we have P(success in longer test) >P(shorter test) but if p is <1/3 , then P(shorter test) >P(longer test). So statement 1 is not sufficient.

Let's see statement 2:

P(Not success in shorter test) >8/9
1-p^2 >8/9
1/9>p^2
p<1/3

So in this case only situation is possible that is P(shorter test success) >P(longer test success)

ALso in question it is just asking which is having better chance.

So 2nd statement is sufficient

So Correct answer is B

Bunuel

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26 Dec 2025, 00:28

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Given that p is the probability of a correct attempt.

The probability of passing the short test is p*p or p^2
The probability of passing the long test is either getting 3 out of 4 correct or all 4 correct.

p^4+ 4C3*p^3(1-p) = p^3[5-4p]

The question is asking whether p^3(5-4p)>p^2

Solving the above equation, we have 4p^2-5p+1<0

By wavy curve method, only possible region where the inequality is satisfied is 1/4<p<1. The question therefore asks whether p is between 1/4 and 1.

Statement 1 - Given here that p^2<1/4 or 0<p<1/2. Not enough since p can be between 0 and 1/4 or 1/4 and 1/2

Statement 2 - Given 1-p^2>8/9 or 0<p<1/3. Not enough since p can be between 0 and 1/3 or 1/3 and 1/4

Combining the statements give us that 0<p<1/3. It is still enough to confirm whether 1/4<p<1.

Therefore, Option E imo

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26 Dec 2025, 00:59

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The probability of qualifying a short test = $p^2$
The probability of qualifying a long test = $4C3 * p^3(1-p) + p^4$, choosing 3 out of 4 and answering 3 questions correctly, one incorrect. And answering all 4 correctly (since it says at least 3 correct).

So the question is asking, if $C^3_4 * p^3(1-p) + p^4 > p^2$?
Simplifying it,
$(3p-1)(p-1) < 0$
Gives, is $p > \frac{1}{3}$? (Note, 0 < p)

(1) $p^2 < \frac{1}{4}$
$\frac{-1}{2}<p<\frac{1}{2}\\
0 $\frac{1}{3}$. Not sufficient

(2) The probability that he qualifies will be less than 1/9.
$p^2 < \frac{1}{9}$
$ 0 < p < \frac{1}{3}$, sufficient to say 'No' to our question.

Option B

Bunuel

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26 Dec 2025, 02:31

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Bunuel