shouldnt the answer here be none ?

A bag contains 3 white and $n$ red marbles, where $n>3$. Two marbles are drawn without replacement.
- Probability that one marble is white and the other is red is greater than
- Probability that both marbles are red.

We need to find all possible integer values of $n$ from the given options for which this inequality holds.

Step 1: Define probabilities
Total marbles $=3+n$

Probability(one white, one red)
This can happen in two ways:
- First white, second red
- First red, second white

So,

$$
\(P(\text { one white }, \text { one red })=P(W R)+P(R W)\)
$$


Calculate each:

$$
\(\begin{aligned}
& P(W R)=\frac{3}{3+n} \times \frac{n}{3+n-1}=\frac{3}{3+n} \times \frac{n}{n+2} \\
& P(R W)=\frac{n}{3+n} \times \frac{3}{3+n-1}=\frac{n}{3+n} \times \frac{3}{n+2}
\end{aligned}\)
$$


Add these:

$$
\(P(\text { one white }, \text { one red })=\frac{3 n}{(3+n)(n+2)}+\frac{3 n}{(3+n)(n+2)}=\frac{6 n}{(3+n)(n+2)}\)
$$

Step 2: Probability(both red)

$$
\(P(\text { both red })=\frac{n}{3+n} \times \frac{n-1}{3+n-1}=\frac{n}{3+n} \times \frac{n-1}{n+2}\)
$$


Step 3: Set inequality

$$
\(\begin{aligned}
& P(\text { one white }, \text { one red })>P(\text { both red }) \\
& \frac{6 n}{(3+n)(n+2)}>\frac{n(n-1)}{(3+n)(n+2)}
\end{aligned}\)
$$


Since denominators are the same and positive for $n>3$, multiply both sides by denominator:

$$
\(6 n>n(n-1)\)
$$


Divide both sides by $n$ (since $n>0$ ):

$$
\(\begin{gathered}
6>n-1 \\
n<7
\end{gathered}\)
$$


Step 4: Check given options $n>3$ and $n<7$
Possible values from options:
- (A) 2 - No, $n>3$ not satisfied.
- (B) 3 - No, $n>3$ not satisfied.
- (C) 4 - Yes, $3<4<7$
- (D) 5 - Yes, $3<5<7$
- (E) 6 - Yes, $3<6<7$

Final answer:
(C) 4, (D) 5, and (E) 6 are the possible values of $n$.