Study the following statements carefully and select the correct option. Cards marked with the consecutive odd numbers from 1to200 are put in a box and mixed thoroughly. One card is drawn at random from the box.
Statemen t - 1 : Probability tha t drawn card is multiple of 3 is`1/3`
Statement· - 2 : Probability that drawn card is a perfect square and a multiple of 9 both is`2/ 3` .

Options

A. Both Statement-1 and Statement-2 are true.
B. Both Staternent-1 and Statement-2 are false.
C. Statement-1 is true but Statement-2 is false.
D. Statement-1 is false but Statement-2 is true.

Answer

To solve the problem, we need to analyze the two statements regarding the probability of drawing a card from a box containing cards marked with consecutive odd numbers from 1 to 200.

### Step 1: Determine the Sample Space
The odd numbers from 1 to 200 are:
1, 3, 5, ..., 199.

To find the total number of odd numbers, we can use the formula for the nth term of an arithmetic progression (AP):
\[ a_n = a + (n-1)d \]
where:
- \( a \) = first term = 1
- \( d \) = common difference = 2
- \( a_n \) = last term = 199

Setting up the equation:
\[ 199 = 1 + (n-1) \cdot 2 \]
\[ 199 - 1 = (n-1) \cdot 2 \]
\[ 198 = (n-1) \cdot 2 \]
\[ n - 1 = \frac{198}{2} = 99 \]
\[ n = 100 \]

So, the total number of cards (sample space) is **100**.

### Step 2: Analyze Statement 1
**Statement 1:** The probability that the drawn card is a multiple of 3 is \( \frac{1}{3} \).

To find the favorable cases (multiples of 3 among the odd numbers):
The multiples of 3 in the odd numbers from 1 to 199 are:
3, 9, 15, ..., 195.

This is also an AP where:
- First term \( a = 3 \)
- Common difference \( d = 6 \) (since we are considering odd multiples of 3)
- Last term \( a_n = 195 \)

Setting up the equation:
\[ 195 = 3 + (n-1) \cdot 6 \]
\[ 195 - 3 = (n-1) \cdot 6 \]
\[ 192 = (n-1) \cdot 6 \]
\[ n - 1 = \frac{192}{6} = 32 \]
\[ n = 33 \]

So, the number of favorable cases (multiples of 3) is **33**.

Now, we calculate the probability:
\[ P(\text{multiple of 3}) = \frac{\text{Number of favorable cases}}{\text{Total cases}} = \frac{33}{100} \]

Since \( \frac{33}{100} \neq \frac{1}{3} \), **Statement 1 is false**.

### Step 3: Analyze Statement 2
**Statement 2:** The probability that the drawn card is a perfect square and a multiple of 9 is \( \frac{2}{3} \).

The perfect squares that are also multiples of 9 among the odd numbers are:
- \( 3^2 = 9 \)
- \( 6^2 = 36 \)
- \( 9^2 = 81 \)
- \( 12^2 = 144 \)

These values are:
9, 36, 81, 144 (all are less than 200).

Counting these, we find there are **4 favorable cases**.

Now, we calculate the probability:
\[ P(\text{perfect square and multiple of 9}) = \frac{4}{100} = \frac{1}{25} \]

Since \( \frac{1}{25} \neq \frac{2}{3} \), **Statement 2 is also false**.

### Conclusion
Both statements are false.

### Final Answer
The correct option is: **Both statement 1 and statement 2 are false.**

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