If X~Bin(n,p) , n=10, then what is value of p if V(X)=2?

To find the value of \( p \) given that \( X \sim \text{Bin}(n, p) \), where \( n = 10 \) and \( V(X) = 2 \), we use the variance formula for a binomial distribution.

The variance \( V(X) \) of a binomial distribution is given by:
\[
V(X) = n p (1 - p)
\]
We are given that \( V(X) = 2 \) and \( n = 10 \). Plugging these values into the equation:
\[
2 = 10 p (1 - p)
\]
Now, simplify and solve for \( p \):
\[
2 = 10p - 10p^2
\]
\[
10p^2 - 10p + 2 = 0
\]
This is a quadratic equation in \( p \). To solve for \( p \), we can use the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 10 \), \( b = -10 \), and \( c = 2 \). Substituting these values:
\[
p = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(10)(2)}}{2(10)}
\]
\[
p = \frac{10 \pm \sqrt{100 - 80}}{20}
\]
\[
p = \frac{10 \pm \sqrt{20}}{20}
\]
\[
p = \frac{10 \pm 4.472}{20}
\]
This gives two possible values for \( p \):
\[
p_1 = \frac{10 + 4.472}{20} = \frac{14.472}{20} = 0.7236
\]
\[
p_2 = \frac{10 - 4.472}{20} = \frac{5.528}{20} = 0.2764
\]

Thus, the two possible values of \( p \) are approximately \( 0.7236 \) and \( 0.2764 \).

Would you like more details on any of the steps? 

Here are five related questions:
1. How do you derive the formula for the variance of a binomial distribution?
2. Can you interpret the meaning of \( p \) in a binomial distribution?
3. What are the implications if \( p \) is closer to 0 or 1 in binomial distributions?
4. How do you solve quadratic equations in general?
5. How does the binomial distribution change as \( n \) increases?

**Tip:** When solving quadratic equations, always check both roots to ensure they make sense in the context of the problem.

Solve for p in a binomial distribution problem where n=10 and V(X)=2 using the variance formula and quadratic equation. Learn key concepts like binomial distributions and quadratic solutions. Suitable for Grades 9-12.

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