For a probability distribution, the sum of the probabilities of all possible outcomes must be equal to 1. In this case, the possible values of X are 0, 1, 2, and 3, and their corresponding probabilities are \( P(X=0) = p \), \( P(X=1) = p \), \( P(X=2) = 0.3 \), and \( P(X=3) = 2p \).
Therefore, we have the equation:
$$ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 $$
$$ p + p + 0.3 + 2p = 1 $$
Combine the terms with \( p \):
$$ 4p + 0.3 = 1 $$
Subtract \( 0.3 \) from both sides:
$$ 4p = 1 - 0.3 $$
$$ 4p = 0.7 $$
Now, solve for \( p \):
$$ p = \frac{0.7}{4} = \frac{7/10}{4} = \frac{7}{10 \times 4} = \frac{7}{40} $$
So, the value of \( p \) is \( \frac{7}{40} \).