The new "extreme" quant question

GRE Quant Problem Solving
1

Given an uniformly-distributed random variable that takes integer values between a and b inclusive.

P(3<=X<=7) = 1/21
P(0<=X<=5) = 1/35

What is P(90<=X<=110) if a<=3 and b>=10?

Shouldn’t the answer be 6/35?

→ Since the distribution is uniform, and 5 integers (3 to 7) contribute 1/21, each integer should contribute 1/105. Also since 3 integers would then contribute 1/35, P(X<=2) = 0. Similar since X takes values between a and b inclusive b should be 107 (to cover total 105 integers)

So P(90<=X<=105) should be P(X=90) + P(X=91) + … +P(X=107) = 18 / 105 = 6/35?

2

You posted this on Reddit as well. We’ve fixed it.

1 Like
3

could you share the reddit link

4

https://www.reddit.com/r/GRE/comments/16k0gac/new_extreme_quant_question_in_gregmat/

5

A Chat GPT Reply:
We are given:

  • A discrete uniform random variable XXX that takes integer values between aaa and bbb, inclusive.
  • That is, X∼U{a,a+1,…,b}X \sim U{a, a+1, \dots, b}X∼U{a,a+1,…,b}
  • So, each integer in that range has equal probability:P(X=x)=1b−a+1,for a≤x≤bP(X = x) = \frac{1}{b - a + 1}, \quad \text{for } a \le x \le bP(X=x)=b−a+11​,for a≤x≤b

Given:

  1. P(3≤X≤7)=121P(3 \le X \le 7) = \frac{1}{21}P(3≤X≤7)=211​
  2. P(0≤X≤5)=135P(0 \le X \le 5) = \frac{1}{35}P(0≤X≤5)=351​
  3. Need to compute P(90≤X≤110)P(90 \le X \le 110)P(90≤X≤110)
  4. We also know that a≤3a \le 3a≤3 and b≥10b \ge 10b≥10

Step 1: General formula for probability

For uniform distribution over integers aaa to bbb:

P(m≤X≤n)=number of integers from m to nb−a+1=n−m+1b−a+1P(m \le X \le n) = \frac{\text{number of integers from } m \text{ to } n}{b - a + 1} = \frac{n - m + 1}{b - a + 1}P(m≤X≤n)=b−a+1number of integers from m to n​=b−a+1n−m+1​

Let’s use this for both of the known probabilities.

Step 2: Use the first equation

P(3≤X≤7)=7−3+1b−a+1=5b−a+1P(3 \le X \le 7) = \frac{7 - 3 + 1}{b - a + 1} = \frac{5}{b - a + 1}P(3≤X≤7)=b−a+17−3+1​=b−a+15​

We are told this equals 121\frac{1}{21}211​, so:

5b−a+1=121⇒b−a+1=105⇒b−a=104\frac{5}{b - a + 1} = \frac{1}{21} \Rightarrow b - a + 1 = 105 \Rightarrow \boxed{b - a = 104}b−a+15​=211​⇒b−a+1=105⇒b−a=104​

Step 3: Use this to compute P(90≤X≤110)P(90 \le X \le 110)P(90≤X≤110)

Now apply the same formula:

P(90≤X≤110)=110−90+1b−a+1=21105=15P(90 \le X \le 110) = \frac{110 - 90 + 1}{b - a + 1} = \frac{21}{105} = \boxed{\frac{1}{5}}P(90≤X≤110)=b−a+1110−90+1​=10521​=51​​

:white_check_mark: Final Answer: 15\boxed{\frac{1}{5}}51​​

6

@user4610 Why?