MATHEMATICS

JAMB 2014 - Question 50

Mathematics 2014 JAMB Past Questions - Question 50: A number is chosen at random from 10 to 30 both inclusive,what is the probability that the number divisible is 3?

A number is chosen at random from 10 to 30 both inclusive,what is the probability that the number divisible is 3?
A:
B:
C:
D:
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Correct Answer

D

Explanation

sample space S = {10,11,12,......30}Let E denote the event of choosing a number divisible by 3.then E = {12,15,18,21,24,27,30} and n(E) = 7therefore prob (E) = n(E) / n(E)To find the probability that a number chosen at random from the range 10 to 30 (inclusive) is divisible by 3, we need to determine the number of favorable outcomes (numbers divisible by 3) and divide it by the total number of possible outcomes.The numbers divisible by 3 in the given range are: 12, 15, 18, 21, 24, 27, and 30.The total number of possible outcomes is the difference between the highest and lowest numbers in the range, plus 1: 30 - 10 + 1 = 21.Therefore, the probability of choosing a number divisible by 3 is 7 (number of favorable outcomes) divided by 21 (total number of possible outcomes):Probability = 7/21 = 1/3So, the probability that the number chosen at random from the range 10 to 30 is divisible by 3 is 1/3.= 7/21=1/3

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