In a workshop there are five machines and the probability of any one of them to be out of service on

Q: Why can't we just use simple probability instead of the Binomial Distribution?

You absolutely could, but it would be tedious! Simple probability works for one or two events, but when you have $5$ machines with independent failure probabilities, the Binomial Distribution is the mathematically elegant way to group these possibilities. It handles the combinations $\binom{n}{r}$ for you, ensuring you don't miss any permutations of which machines are faulty.

Q: Wait, why do we factor out $(\frac{3}{4})^3$? Can't I just calculate each term individually?

You can, but in JEE Advanced, time is your most precious resource. Calculating $(\frac{3}{4})^5$ or $(\frac{3}{4})^4$ involves larger numbers and higher chances of arithmetic errors. Factoring out the common term $(\frac{3}{4})^3$ reduces the problem to simple fraction addition inside the bracket, which is much faster and safer.

Q: Is there a trick if the question asked for 'at least two' instead of 'at most two'?

Great question! If it were 'at least two', you would calculate $P(X \ge 2) = 1 - [P(X=0) + P(X=1)]$. Always look for the complement if the number of cases to calculate is large. Here, 'at most two' ($0, 1, 2$) is three cases, while 'at least two' ($2, 3, 4, 5$) is four cases. Always choose the path with fewer calculations!

Question

In a workshop there are five machines and the probability of any one of them to be out of service on a day is $1/4$. If the probability that at most two machines will be out of service on the same day is $(3/4)^3 * k$ then $k$ is equal to:

Options:

(A) 17/2

(B) 4

(C) 17/8

(D) 17/4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the manager of a high-tech workshop with five machines. Each machine is temperamental, with the probability of being out of service given by $p = \frac{1}{4}$.

This implies the probability of a machine working perfectly is $q = 1 - p = \frac{3}{4}$. Because each machine operates independently, we utilize the **Binomial Distribution**.

Let $X$ be the number of machines out of service. With $n=5$ trials, $X$ follows the distribution $X \sim B(n, p)$. Our goal is to find the probability that 'at most two' machines are out of service, expressed as $P(X \le 2)$.

### Breaking Down the 'At Most' Barrier

The condition 'at most two' means we are interested in the scenarios where zero, one, or exactly two machines are broken. Anything beyond that—three, four, or five—is outside our target event.

The total probability is the sum of these three mutually exclusive scenarios:

$$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$$

We use the Binomial probability mass function: $P(X = r) = \binom{n}{r} p^r q^{n-r}$. Let us calculate each component.

### The Three Pillars of Probability

For $X=0$ (zero machines out):

$$P(X=0) = \binom{5}{0} \left(\frac{1}{4}ight)^0 \left(\frac{3}{4}ight)^5 = 1 	imes 1 	imes \left(\frac{3}{4}ight)^5 = \left(\frac{3}{4}ight)^5$$

For $X=1$ (one machine out):

$$P(X=1) = \binom{5}{1} \left(\frac{1}{4}ight)^1 \left(\frac{3}{4}ight)^4 = 5 	imes \frac{1}{4} 	imes \left(\frac{3}{4}ight)^4$$

For $X=2$ (two machines out):

$$P(X=2) = \binom{5}{2} \left(\frac{1}{4}ight)^2 \left(\frac{3}{4}ight)^3 = 10 	imes \frac{1}{16} 	imes \left(\frac{3}{4}ight)^3$$

### The Elegant Simplification

To sum these, we observe that $(3/4)^5$, $(3/4)^4$, and $(3/4)^3$ all share a common factor of $(3/4)^3$. Factoring this out simplifies the arithmetic significantly:

$$P = \left(\frac{3}{4}ight)^3 \left[ \left(\frac{3}{4}ight)^2 + 5\left(\frac{1}{4}ight)\left(\frac{3}{4}ight) + 10\left(\frac{1}{16}ight) ight]$$

Inside the bracket, we calculate the sum:

$$\frac{9}{16} + \frac{15}{16} + \frac{10}{16} = \frac{34}{16} = \frac{17}{8}$$

### The Final Reveal

We are left with the expression $P = \left(\frac{3}{4}ight)^3 	imes \frac{17}{8}$. Given that this probability is defined as $\left(\frac{3}{4}ight)^3 	imes k$, we compare the two sides to find $k$.

The value of $k$ is:

**$$k = \frac{17}{8}$$**

By navigating the logic of independent events and using algebraic factoring, we have bypassed tedious arithmetic. Always look for the structure of the expression before performing final calculations to ensure accuracy in your JEE solutions.

Animated Solution for Mathematics - Probability: In a workshop there are five machines and the probability of any one of them to be out of service on a day is $1/4$ . If the probability that at most two machines will be out of service on the same day is $(3/4)^{3} * k$ then $k$ is equal to:

Select Answer:

Visualized Solution

Understanding the Scenario

The Binomial Formula

Defining 'At Most Two'

Case 1: Zero Machines Out ( $X = 0$ )

Case 2: One Machine Out ( $X = 1$ )

Case 3: Two Machines Out ( $X = 2$ )

Summing the Probabilities

Factoring Out $(\frac{3}{4})^{3}$

Simplifying Inside the Bracket

Final Calculation of the Bracket

Finding the Value of $k$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Breaking Down the 'At Most' Barrier

The Three Pillars of Probability

The Elegant Simplification

The Final Reveal

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