Mathematics 2013 JAMB Past Questions - Question 49: What is the probability that an integer x(1 ≤ x ≤ 25) chosen at random is divisible by both 2 and 3?

Correct Answer

A

Explanation

(1 < x < 25) = {1,2,3,4,5,,(6),7,8,9,10,11,(12),13,14,15,16,17,(18), 19,20, 21,22,23,(24),25}number Nof x divisible by both 2 and 3 is 4. n(e) = 25.Hence the required probability = N / n(e) = 4/25To find the probability that an integer x (1 ≤ x ≤ 25) chosen at random is divisible by both 2 and 3, we need to determine the number of integers in the given range that satisfy this condition and divide it by the total number of integers in the range.First, let's find the number of integers divisible by 2 in the range 1 to 25. We can observe that every other integer in this range is divisible by 2. So, there are 25/2 = 12 integers divisible by 2.Next, let's find the number of integers divisible by 3 in the range 1 to 25. We can observe that every third integer in this range is divisible by 3. So, there are 25/3 = 8 integers divisible by 3.To find the number of integers divisible by both 2 and 3, we need to find the common multiples of 2 and 3 in the range 1 to 25. These common multiples are 6, 12, 18, and 24. So, there are 4 integers divisible by both 2 and 3.Therefore, the probability that an integer x chosen at random from the range 1 to 25 is divisible by both 2 and 3 is given by:Probability = (Number of integers divisible by both 2 and 3) / (Total number of integers in the range)Probability = 4 / 25Simplifying, we get:Probability = 0.16 or 16%Therefore, the probability that an integer x chosen at random from the range 1 to 25 is divisible by both 2 and 3 is 16%.