MCQ
A three digit number is chosen at random, the probability that it is divisible by both 2 and 3, is
- A$\frac{1}{8}$
- B$\frac{1}{9}$
- ✓$\frac{1}{6}$
- D$\frac{1}{12}$
✓
Answer
Correct option: C.
$\frac{1}{6}$
(c) Three digit numbers are. 100, 101, ..., 999. These are 900 numbers out of which one can be chosen in 900 ways. So, total number of elementary events is 900.
A number will be divisible by both 2 and 3, if it is divisible by 6. Three digit numbers divisible by 6 are 102, 108, 114, ...., 996. These numbers form an AP with first term = 102, common difference d = 6 and last term = 996. Let there number be n. Then, $996=102+(n-1) \times 6 \Rightarrow 894=6(n-1) \Rightarrow n-1=149 \Rightarrow n=150$
So, there are 150 numbers between 102 and 996 which are divisible by 6. Out of these one number can be chosen in 150 ways.
Hence, required probability $=\frac{150}{900}=\frac{1}{6}$.
A number will be divisible by both 2 and 3, if it is divisible by 6. Three digit numbers divisible by 6 are 102, 108, 114, ...., 996. These numbers form an AP with first term = 102, common difference d = 6 and last term = 996. Let there number be n. Then, $996=102+(n-1) \times 6 \Rightarrow 894=6(n-1) \Rightarrow n-1=149 \Rightarrow n=150$
So, there are 150 numbers between 102 and 996 which are divisible by 6. Out of these one number can be chosen in 150 ways.
Hence, required probability $=\frac{150}{900}=\frac{1}{6}$.
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