Probability Question 45
Question: A three digit number is formed by using numbers 1, 2, 3 and 4. The probability that the number is divisible by 3, is
Options:
A) $ \frac{2}{3} $
B) $ \frac{2}{7} $
C) $ \frac{1}{2} $
D) $ \frac{3}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
Total number of ways to form the numbers of three digit with 1, 2, 3 and 4 are $ {}^{4}P_3=4!=24 $
If the numbers are divisible by three then their sum of digits must be 3, 6 or 9
But sum 3 is impossible.
Then for sum 6, digits are 1, 2, 3 Number of ways $ =3! $
Similarly for sum 9, digits are 2, 3, 4.
Number of ways =3!
Thus number of favorable ways $ =3!+3! $
Hence required probability $ =\frac{3!+3!}{4!}=\frac{12}{24}=\frac{1}{2}. $
BETA
