Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?
\(\frac{2}{3}\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
Explanation
Video Explanation
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Discussions (13)
You can't have 4,4 or 3,3 or 2,2 or 1,1 as possibilities because the question clearly states "Two numbers are removed at random from the numbers 1, 2, 3 and 4"
I think the question is pretty straightforward. There are only six possibilities given just those four numbers : 1,2; 1,3; 1,4; 2,4; 2,3; 3,4. Just two even sum so the answer is 1/3 (C).
The man who solved the question in the video, mentioned it that we were not told that there was repetition in picking the two numbers, still, he decided to solve the question using the repetition method!‽‽‽. These people ehn 🤦😤.
I pray they make corrections to this on time before people are misled.
The question didn't mention any repetition or two same numbers. Which means there are only 2 possible scenarios where the 2 randomly picked numbers summed up are even. Which equates to 2/6 = ⅓. Why Myschool is picking Option B. those students that don't really know or understand the topic well will be misled now. why this rubbish?
and this is not the first time. I've met a lot of questions where Myschool provided awfully wrong answers, while some incorrect explanations
To solve this quickly, you don't need to list every possible outcome. You just need to look at the parity (odd/even nature) of the numbers.
The Fast Logic
For the sum of two numbers to be even, they must either both be odd or both be even.
* Identify the groups: In the set \{1, 2, 3, 4\}, there are:
* 2 Odd numbers: \{1, 3\}
* 2 Even numbers: \{2, 4\}
* Count the "Even Sum" combinations:
* Pair of Odds: (1, 3) — 1 way
* Pair of Evens: (2, 4) — 1 way
* Total favorable outcomes = 2
* Count total possible pairs:
From 4 numbers, the total number of ways to pick 2 is calculated as:
(Or simply list them mentally: 1-2, 1-3, 1-4, 2-3, 2-4, 3-4)
The Calculation
Wait, why does the image show 1/2 as correct?
Looking at your screenshot, the system marked B (1/2) as the correct answer and C (1/3) as wrong.
Strictly speaking, mathematically, 1/3 is the correct answer for selecting two distinct numbers from that set. The only way the answer could be 1/2 is if the numbers were removed one at a time with replacement (meaning you could pick the same number twice, like 1 and 1), but the phrasing "removed from the numbers" almost always implies selection without replacement.
If this is for a specific standardized test, they might have an error in their answer key, or they are assuming the order of removal matters (which it shouldn't for a sum). Stick with the 1/3 logic for any formal math setting!
But they never talked about repetition. So if there is no repetition, it should have only 6 possibilities ( ͡°_ʖ ͡°)
the answer is 1/2
we have 2 even numbers in the numbers given, so we say 2/4 because there are 4 total numbers in the numbers (1,2,3,4) say 2/4 1/2
