Suppose A and B are two events and P(B) = p, P(A) = `p^2` where 0 `lt`p `lt` 1. If the odds against A are the cubes of the odds against B, then 6p- 1.2 is equal to ______

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the Odds
The odds against an event A is defined as the ratio of the probability of the event not occurring to the probability of the event occurring. Mathematically, this can be expressed as:
\[ \text{Odds against A} = \frac{P(A')}{P(A)} = \frac{1 - P(A)}{P(A)} \]
where \( P(A') \) is the probability of A not occurring.

### Step 2: Define the Probabilities
Given:
- \( P(B) = p \)
- \( P(A) = p^2 \)

Thus, we can express the odds against A and B:
\[ \text{Odds against A} = \frac{1 - p^2}{p^2} \]
\[ \text{Odds against B} = \frac{1 - p}{p} \]

### Step 3: Set Up the Relationship
According to the problem, the odds against A are the cubes of the odds against B:
\[ \frac{1 - p^2}{p^2} = \left(\frac{1 - p}{p}\right)^3 \]

### Step 4: Simplify the Equation
Cross-multiplying gives:
\[ (1 - p^2) p^3 = (1 - p)^3 \]

### Step 5: Expand Both Sides
Expanding the right side using the binomial theorem:
\[ (1 - p)^3 = 1 - 3p + 3p^2 - p^3 \]
Thus, we have:
\[ p^3 - p^5 = 1 - 3p + 3p^2 - p^3 \]

### Step 6: Rearranging the Equation
Rearranging the equation gives:
\[ p^5 - 2p^3 + 3p - 1 = 0 \]

### Step 7: Factor the Polynomial
We can factor this polynomial. Notice that \( p = 1 \) is not a valid solution since \( 0 < p < 1 \). We can try \( p = \frac{1}{3} \):
\[ 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right) - 1 = 0 \]
This simplifies to:
\[ \frac{3}{9} - \frac{2}{27} + 1 - 1 = 0 \]
Thus, \( p = \frac{1}{3} \) is a valid solution.

### Step 8: Calculate \( 6p - 1.2 \)
Now substituting \( p = \frac{1}{3} \):
\[ 6p - 1.2 = 6 \times \frac{1}{3} - 1.2 = 2 - 1.2 = 0.8 \]

### Final Answer
Thus, the value of \( 6p - 1.2 \) is:
\[ \boxed{0.8} \]