Understanding Binomial Distribution Mean and Variance

The Binomial distribution describes the number of successes in a fixed number of independent trials, each having two possible outcomes (success or failure). Key parameters are:

• Number of trials ($n$): The fixed number of times the experiment is conducted.
• Probability of success ($p$): The probability of achieving success in a single trial.

The mean ($\mu$) and variance ($\sigma^2$) of a Binomial distribution are given by:

• Mean: $\mu = np$
• Variance: $\sigma^2 = npq$, where $q = 1-p$ (probability of failure).

Calculating Probability Parameters

We are given that the sum of the mean and variance is 1.8:

$ \mu + \sigma^2 = 1.8 $

Substituting the formulas for mean and variance:

$ np + npq = 1.8 $

We know the number of trials is $n=5$. Substitute this value:

$ 5p + 5pq = 1.8 $

Since $q = 1-p$, substitute this into the equation:

$ 5p + 5p(1-p) = 1.8 $

Now, simplify and solve for $p$:

$ 5p + 5p - 5p^2 = 1.8 $

$ 10p - 5p^2 = 1.8 $

Rearrange into a standard quadratic equation ($ax^2 + bx + c = 0$):

$ 5p^2 - 10p + 1.8 = 0 $

Use the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=5$, $b=-10$, $c=1.8$:

$ p = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(5)(1.8)}}{2(5)} $

$ p = \frac{10 \pm \sqrt{100 - 36}}{10} $

$ p = \frac{10 \pm \sqrt{64}}{10} $

$ p = \frac{10 \pm 8}{10} $

This gives two possible values for $p$:

• $ p_1 = \frac{10 + 8}{10} = \frac{18}{10} = 1.8 $ (This is not a valid probability as it's greater than 1)
• $ p_2 = \frac{10 - 8}{10} = \frac{2}{10} = 0.2 $ (This is a valid probability)

Therefore, the probability of success is $p=0.2$.

We can calculate the probability of failure, $q$:

$ q = 1 - p = 1 - 0.2 = 0.8 $

Calculating Probability of Two Successes

The question asks for the probability of exactly two successes ($k=2$) in $n=5$ trials.

The Binomial probability formula is:

$ P(X=k) = \binom{n}{k} p^k q^{n-k} $

Substitute the values $n=5$, $k=2$, $p=0.2$, and $q=0.8$:

$ P(X=2) = \binom{5}{2} (0.2)^2 (0.8)^{5-2} $

$ P(X=2) = \binom{5}{2} (0.2)^2 (0.8)^3 $

First, calculate the binomial coefficient $\binom{5}{2}$:

$ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 $

Next, calculate the powers:

• $ (0.2)^2 = 0.04 $
• $ (0.8)^3 = 0.8 \times 0.8 \times 0.8 = 0.512 $

Finally, multiply these values together:

$ P(X=2) = 10 \times 0.04 \times 0.512 $

$ P(X=2) = 0.4 \times 0.512 $

$ P(X=2) = 0.2048 $

Thus, the probability of achieving exactly two successes is 0.2048.