MCQ
A three digit number is chosen at random, the probability that its hundred's digit, ten's digit and unit's digit are consecutive integers in descending order, is
- A$\frac{1}{75}$
- B$\frac{4}{225}$
- ✓$\frac{2}{225}$
- D$\frac{1}{45}$
✓
Answer
Correct option: C.
$\frac{2}{225}$
(c) There are 900 three digit numbers, namely, 100, 101, ..., 998, 999. Out of these one number can be chosen in 900 ways.
∴ Total number of elementary events = 900.
Three digit numbers having hundred's, ten's and unit's digit as consecutive integers in descending order are 210, 321, 432, 543, 654, 765, 876, 987. These are 8 numbers, out of which one can be chosen in 8 ways.
∴ Favourable number of elementary events = 8
Hence, required probability =$\frac{8}{900}=\frac{2}{225}$.
∴ Total number of elementary events = 900.
Three digit numbers having hundred's, ten's and unit's digit as consecutive integers in descending order are 210, 321, 432, 543, 654, 765, 876, 987. These are 8 numbers, out of which one can be chosen in 8 ways.
∴ Favourable number of elementary events = 8
Hence, required probability =$\frac{8}{900}=\frac{2}{225}$.
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