A box contains card bearing numbers from 6 to 70. If one card is drawn at random from the box, find the porbability that it bears (i) a one digit number(ii) a number divisible by 5.

Answer

To solve the problem, we need to find the probability of drawing a card from a box containing numbers from 6 to 70 under two conditions: (i) the card bears a one-digit number and (ii) the card bears a number divisible by 5.

### Step-by-Step Solution:

1. **Determine the Total Number of Cards:**
The cards are numbered from 6 to 70. To find the total number of cards, we can use the formula for the number of terms in an arithmetic progression (AP):
\[ n = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 \]
Here, the first term (a) is 6, the last term (l) is 70, and the common difference (d) is 1.
\[ n = \frac{70 - 6}{1} + 1 = 64 + 1 = 65 \]
Thus, there are **65 cards** in total.

2. **Find the Probability of Drawing a One-Digit Number:**
The one-digit numbers in the range from 6 to 70 are 6, 7, 8, and 9.
- Count the one-digit numbers: 6, 7, 8, 9 (Total = 4).
- The probability \( P(A) \) of drawing a one-digit number is given by:
\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{65} \]

3. **Find the Probability of Drawing a Number Divisible by 5:**
The numbers between 6 and 70 that are divisible by 5 are: 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, and 70.
- Count these numbers: 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70 (Total = 13).
- The probability \( P(B) \) of drawing a number divisible by 5 is given by:
\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{13}{65} = \frac{1}{5} \]

### Final Answers:
- (i) The probability that the card bears a one-digit number is \( \frac{4}{65} \).
- (ii) The probability that the card bears a number divisible by 5 is \( \frac{1}{5} \).