The range of a random variable X is {0, 1, 2}. Given that `P(X=0)=3c^(3),P(X=1)=4c-10c^(2),P(X=2)=5c-1` where c is constant.
Find (i) the value of c (ii) `P(X lt 1)`
(iii) `P(1lt X le2)` (iv) `P(0lt X le3)`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript.

### Step 1: Find the value of \( c \)

We know that the sum of the probabilities must equal 1:

\[ P(X=0) + P(X=1) + P(X=2) = 1 \]

Substituting the given probabilities:

\[ 3c^3 + (4c - 10c^2) + (5c - 1) = 1 \]

Combining like terms:

\[ 3c^3 - 10c^2 + 9c - 2 = 0 \]

Now we need to find the roots of this cubic equation. We can test possible rational roots.

Testing \( c = 1 \):

\[ 3(1)^3 - 10(1)^2 + 9(1) - 2 = 3 - 10 + 9 - 2 = 0 \]

Thus, \( c = 1 \) is a root. We can factor the cubic polynomial using \( c - 1 \).

Using synthetic division or polynomial long division, we divide \( 3c^3 - 10c^2 + 9c - 2 \) by \( c - 1 \):

The result of the division is:

\[ 3c^2 - 7c + 2 \]

Now we can find the roots of the quadratic equation \( 3c^2 - 7c + 2 = 0 \) using the quadratic formula:

\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting \( a = 3, b = -7, c = 2 \):

\[ c = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \]

Calculating the two possible values:

1. \( c = \frac{12}{6} = 2 \)
2. \( c = \frac{2}{6} = \frac{1}{3} \)

Now we need to check which values of \( c \) give valid probabilities.

- For \( c = 1 \): \( P(X=1) = 4(1) - 10(1^2) = 4 - 10 = -6 \) (not valid)
- For \( c = 2 \): \( P(X=1) = 4(2) - 10(2^2) = 8 - 40 = -32 \) (not valid)
- For \( c = \frac{1}{3} \):

\[ P(X=1) = 4\left(\frac{1}{3}\right) - 10\left(\frac{1}{3}\right)^2 = \frac{4}{3} - \frac{10}{9} = \frac{12}{9} - \frac{10}{9} = \frac{2}{9} \quad (\text{valid}) \]

Thus, the only valid value for \( c \) is:

\[ \boxed{\frac{1}{3}} \]

### Step 2: Find \( P(X < 1) \)

Since \( P(X < 1) = P(X = 0) \):

\[ P(X=0) = 3c^3 = 3\left(\frac{1}{3}\right)^3 = 3 \cdot \frac{1}{27} = \frac{1}{9} \]

Thus,

\[ \boxed{\frac{1}{9}} \]

### Step 3: Find \( P(1 < X \leq 2) \)

This is equal to \( P(X = 2) \):

\[ P(X=2) = 5c - 1 = 5\left(\frac{1}{3}\right) - 1 = \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]

Thus,

\[ \boxed{\frac{2}{3}} \]

### Step 4: Find \( P(0 < X \leq 3) \)

Since \( X \) can only take values 0, 1, and 2, we have:

\[ P(0 < X \leq 3) = P(X=1) + P(X=2) \]

Calculating \( P(X=1) \):

\[ P(X=1) = 4c - 10c^2 = 4\left(\frac{1}{3}\right) - 10\left(\frac{1}{3}\right)^2 = \frac{4}{3} - \frac{10}{9} = \frac{12}{9} - \frac{10}{9} = \frac{2}{9} \]

Now adding \( P(X=2) \):

\[ P(0 < X \leq 3) = P(X=1) + P(X=2) = \frac{2}{9} + \frac{2}{3} = \frac{2}{9} + \frac{6}{9} = \frac{8}{9} \]

Thus,

\[ \boxed{\frac{8}{9}} \]

### Summary of Answers
1. \( c = \frac{1}{3} \)
2. \( P(X < 1) = \frac{1}{9} \)
3. \( P(1 < X \leq 2) = \frac{2}{3} \)
4. \( P(0 < X \leq 3) = \frac{8}{9} \)