Jambmaths question:
What is the probability that an integer x $(1\le x\le 20)$chosen at random is divisible by both 2 and 3
Option A:
$\tfrac{3}{20}$
Option B:
$\tfrac{7}{10}$
Option C:
$\tfrac{1}{20}$
Option D:
$\tfrac{1}{3}$
Jamb Maths Solution:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18.19,20
Total possible outcome =20
Number divisible by both 2 and 3 are 6,12, 18
Number of expected outcome = 3
$Probability=\frac{number of expected outcome}{number of total possible outcome}$
$Probability(\text{ divisible by 2 and 3)}=\frac{3}{20}$
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