A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain fixed number N (N ≥ 2) o

Q: Why can't we just calculate the probability directly? It seems like it would just be a few cases.

Ah, the classic trap! If you try to calculate 'less than 150' directly, you have to account for every combination of 50p, 25p, and the N coins that sum to 145, 140, 135, and so on. It is a combinatorial explosion. By using the complement, we reduce the problem to just three specific scenarios where the sum is high. Always look for the path of least resistance.

Q: Wait, why do we treat the 10p and 5p coins as just 'N' coins? Doesn't their value matter?

Great observation! In the context of the complementary event (sum $\ge 150$), we only care if the sum hits the threshold. Since even the largest of these 'N' coins (10p) is small, they act as 'fillers' in our cases. We don't need to distinguish between 10p and 5p because, in the specific case where we need them to reach 150, they are simply 'one coin from the N pool'. The math holds regardless of their specific value.

Q: How do I know I haven't missed a case in the complement?

The beauty of the 'highest denomination' strategy is that it is exhaustive. We started with the maximum possible 50p coins (2) and worked down. Once we reached a point where even the maximum possible combination of remaining coins couldn't reach 150 (like 0 fifty-paise coins), we knew we had covered all possibilities. Trust your systematic partitioning!

Question

A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain fixed number $N (N \ge 2)$ of ten and five paise coins. Five coins are taken out of the box at random. Find the probability that the total value of these 5 coins is less than one rupee and fifty paise.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a probability problem; we are learning how to look at a chaotic system and impose order upon it.

Imagine you are standing before a box filled with coins: 2 fifty-paise coins, 5 twenty-five-paise coins, and $N$ smaller coins (10p or 5p). You are asked to reach in and grab 5 coins at random.

The question is: what is the probability that their total value is less than 150 paise? If you try to count every single combination that sums to less than 150, you will be here until the next JEE exam.

This is where we pivot. We use the power of the **Complementary Event**. Instead of counting the 'less than' scenarios, we count the 'greater than or equal to' scenarios and subtract that from 1.

### Defining the Universe

First, we must define our sample space. We have 2 fifty-paise coins, 5 twenty-five-paise coins, and $N$ smaller coins. The total number of coins is $2 + 5 + N = N + 7$.

We are choosing 5 coins from this total. The total number of ways to do this is given by the combination formula:

$$
\binom{N+7}{5}
$$

This is our denominator, the universe of all possible outcomes.

### The Strategic Pivot

We define our target event $E$ as the total value being less than 150 paise. The complement, $E^c$, is the event where the total value is $\ge 150$ paise.

Our strategy is simple:

$$
P(E) = 1 - P(E^c)
$$

Now, how do we find $P(E^c)$? We need to reach a sum of at least 150 paise using only 5 coins. To do this, we must be greedy and prioritize the highest denomination coins: the 50p coins.

### The Case Analysis

Let's break this down systematically.

**Case 1:** We select both 50p coins. That gives us 100 paise. We need 3 more coins to reach 5 total. If we pick 3 twenty-five-paise coins, we get $100 + 3(25) = 175$ paise. This is $\ge 150$. The number of ways is:

$$
\binom{2}{2} 	imes \binom{5}{3} = 1 	imes 10 = 10
$$

**Case 2:** We select both 50p coins (100 paise) and only 2 twenty-five-paise coins (50 paise). We need 1 more coin from the $N$ pool. The total is $100 + 50 + (10 	ext{ or } 5) = 160$ or $155$ paise. Both are $\ge 150$. The number of ways is:

$$
\binom{2}{2} 	imes \binom{5}{2} 	imes \binom{N}{1} = 10N
$$

**Case 3:** We select only 1 fifty-paise coin (50 paise). To reach 150, we need 100 more paise from the remaining 4 coins. The only way is to pick 4 twenty-five-paise coins. Total: $50 + 4(25) = 150$ paise. This is exactly 150, which satisfies our condition. The number of ways is:

$$
\binom{2}{1} 	imes \binom{5}{4} = 2 	imes 5 = 10
$$

### The Synthesis

We have exhausted all possibilities. If we pick 0 fifty-paise coins, the maximum value is $5(25) = 125$, which is less than 150.

The total favorable ways for $E^c$ is $10 + 10N + 10 = 10(N+2)$. Finally, the probability of the complement is:

$$
P(E^c) = \frac{10(N+2)}{\binom{N+7}{5}}
$$

Subtracting this from 1 gives us our **final answer**:

$$
P(E) = 1 - \frac{10(N+2)}{\binom{N+7}{5}}
$$

See how elegant that is? By breaking the problem into cases and using the complement, we turned a mountain of complexity into a simple, manageable calculation. Keep this mindset, and no problem will ever be too difficult for you.

Visualized Solution

Inventory of the Box

The Complementary Strategy

Analyzing the Complementary Event

Case 1: Two 50p and Three 25p Coins

Case 2: Two 50p, Two 25p, and One N Coin

Case 3: One 50p and Four 25p Coins

Checking Other Possibilities

Total Favorable Ways for $E^{c}$

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Defining the Universe

The Strategic Pivot

The Case Analysis

The Synthesis

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