Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2

Q: Wait, why do we divide the two equations? Can't we just solve them like a normal system of equations?

You absolutely can! However, in competitive exams like JEE, time is your most precious resource. Dividing the equations is a strategic 'shortcut' because it causes the binomial coefficients and the variables to cancel out partially, leading you to a simple linear relationship between $p$ and $q$ much faster than standard substitution would.

Q: I keep getting confused with the binomial coefficient $\binom{n}{k}$. Is there a trick to remember it?

Think of $\binom{n}{k}$ as the number of ways to choose $k$ successful outcomes out of $n$ total trials. For $n=5$, you can use Pascal's Triangle: the row for $n=5$ is $1, 5, 10, 10, 5, 1$. So, $\binom{5}{1}=5$ and $\binom{5}{2}=10$. Visualizing this triangle is often faster than calculating the factorials manually!

Q: Why is the probability of failure $q$ equal to $1-p$?

This is the fundamental axiom of probability for a binary outcome. In any single trial, you either succeed or you fail. There is no third option. Therefore, the sum of the probability of success ($p$) and the probability of failure ($q$) must account for the entire sample space, which is $1$ (or $100\%$).

Question

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:

Options:

(A) $32/625$

(B) $80/243$

(C) $40/243$

(D) $128/625$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are conducting an experiment. You repeat it five times, and each time, the outcome is binary: either a success or a failure. This is the classic setup for a Binomial Distribution.

In this problem, we are given the probabilities of exactly one success and exactly two successes, and we are tasked with finding the probability of exactly three. It might seem like a daunting algebraic puzzle, but let us break it down into a logical, elegant narrative.

### The Engine of Our Calculation

To model this, we rely on the Binomial Distribution formula:

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

Here, $n=5$ is the number of trials, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure. This formula is the engine of our solution.

We are given $P(X=1) = 0.4096$ and $P(X=2) = 0.2048$. Our first step is to translate these into mathematical equations:

For $k=1$:
$$P(X=1) = \binom{5}{1} p^1 q^4 = 5pq^4 = 0.4096$$

For $k=2$:
$$P(X=2) = \binom{5}{2} p^2 q^3 = 10p^2q^3 = 0.2048$$

### The JEE Secret Weapon: The Ratio Method

Now, we face a system of two equations. By taking the ratio of the two equations, we can make the variables dance to our tune. Let us divide the second equation by the first:

$$\frac{P(X=2)}{P(X=1)} = \frac{10p^2q^3}{5pq^4}$$

Look at how beautifully the terms simplify! The coefficients $10$ and $5$ reduce to $2$. The $p^2$ and $p$ reduce to $p$ in the numerator, and the $q^3$ and $q^4$ reduce to $q$ in the denominator.

We are left with:

$$\frac{2p}{q} = \frac{0.2048}{0.4096}$$

Since $0.2048$ is exactly half of $0.4096$, the right side simplifies to $\frac{1}{2}$. Thus, we have $\frac{2p}{q} = \frac{1}{2}$, which leads us to the clean, powerful relationship:

$$q = 4p$$

### Solving the Core Parameters

We are now standing on the threshold of the solution. We know that $p + q = 1$.

By substituting our new relationship $q = 4p$ into this fundamental constraint, we get $p + 4p = 1$, which means $5p = 1$, or $p = 0.2$. Consequently, $q = 0.8$.

We have successfully decoded the DNA of this binomial distribution. We know exactly how likely success and failure are in each trial.

### The Grand Finale

With $p = 0.2$ and $q = 0.8$ in hand, finding the probability of exactly three successes is just a matter of plugging the values into our formula. We need $P(X=3) = \binom{5}{3} p^3 q^2$.

We know $\binom{5}{3} = 10$. So:

$$P(X=3) = 10 	imes (0.2)^3 	imes (0.8)^2$$

Calculating this, we get:

$$P(X=3) = 10 	imes 0.008 	imes 0.64 = 0.0512$$

Converting this to a fraction, we find $0.0512 = \frac{512}{10000} = \frac{32}{625}$.

We have arrived at our destination. The final answer is **$\frac{32}{625}$** (or **$0.0512$**).

Animated Solution for Mathematics - Probability: Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:

Select Answer:

Visualized Solution

Problem Setup

The Binomial Distribution Formula

Equation for $P (X = 1)$

Simplifying $P (X = 1)$

Equation for $P (X = 2)$

Simplifying $P (X = 2)$

The Ratio Method

Simplifying the Ratio

Relation between $p$ and $q$

Using the Probability Constraint

Solving for $p$ and $q$

Setting up for $P (X = 3)$

Final Substitution

Final Calculation

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Engine of Our Calculation

The JEE Secret Weapon

Solving the Core Parameters

The Grand Finale

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