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**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?**

**A.**** 2****/****3**

**B.**** 1****/****2**

**C.**** 1****/****3**

**D.**** 1****/****4**

**QUICK ANSWER…**

**C**

**DETAILS…**

**If t****wo numbers are removed at random from the numbers 1, 2, 3****, ****and 4****, ****the probability that the sum of the numbers removed is even**** ****will be = 1/3**

**Removing two numbers gives:**

**(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) = 6 combinations.**

**The number combinations that will add up to an even number will be**

**= (1, 3) and (2, 4) = 2 combinations.**

**So, ****the probability that the sum of the numbers removed is even**** ****will be = 2/6 = 1/3**

**Now for the right answer to the above question:**

**Option A is incorrect.****Option B is not correct.****C is correct.****D is not the correct answer.**

**KEY-POINTS…**

**You may please note these/this:**

**Probability = number of required combination /number of possible combinations.**

**If you love our answers, you can simply join our community and also provide answers like this, fellow learners like you will appreciate it.**

**/ culled from 20****21**** JAMB-UTME**** MATHEMATICS**** question**** 24 ****/**