Use the probability density function given below to find the indicated probabilities. \[ f(x)=\left\{\begin{array}{cc} \frac{1}{8} x & \text { if } 0 \leq x \leq 4 \\ 0 & \text { otherwise } \end{array}\right. \] (A) $P(X=2)$ (B) $P(X>6)$ (C) $P(X<6)$ (A) $P(X=2)=\square$ (Type an integer or a simplified fraction.) (B) $P(X>6)=\square$ (Type an integer or a simplified fraction.) (C) $P(X<6)=\square$ (Type an integer or a simplified fraction.)

Step 1 ：The problem is asking for the probability of a continuous random variable at a specific point, the probability of a continuous random variable being greater than a certain value, and the probability of a continuous random variable being less than a certain value.

Step 2 ：For a continuous random variable, the probability at a specific point is always 0. This is because the probability is defined as the area under the curve of the probability density function, and a single point has no area. Therefore, \(P(X=2)=\boxed{0}\).

Step 3 ：For the probability of a continuous random variable being greater than a certain value, we need to integrate the probability density function from that value to infinity. However, in this case, the probability density function is 0 for all values greater than 4, so the probability of the random variable being greater than 6 is 0. Therefore, \(P(X>6)=\boxed{0}\).

Step 4 ：For the probability of a continuous random variable being less than a certain value, we need to integrate the probability density function from negative infinity to that value. In this case, the probability density function is 0 for all values less than 0, so we only need to integrate from 0 to 6. However, since the probability density function is 0 for all values greater than 4, we only need to integrate from 0 to 4. Therefore, \(P(X<6)=\boxed{1}\).