Quality Control, Defective Batches, and Geometric Probability for Class 10

In modern industrial manufacturing—from microchips and pharmaceutical capsules to electric LED bulbs and automobile ball bearings—factories produce millions of units daily. Quality control inspectors cannot test every single light bulb before packing, because testing to destruction would leave no products to sell! Instead, engineers draw random quality-control inspection samples to mathematically calculate the probability of defective units.

In CBSE Class 10 Mathematics, Chapter 14 (Probability), defective batch problems, numbered disc questions, and continuous geometric probability represent high-weightage 3-mark and 4-mark board examination questions in Section C and Section D.

What You Will Learn

• Quality control inspection models: Good vs. Defective items
• Non-Replacement Problems: How the sample space shrinks when items are kept aside
• The 90 Numbered Discs Problem (Two-digit numbers, perfect squares, divisibility)
• What is Geometric Probability? Continuous 2D area probability models
• The famous NCERT Helicopter Crash into a Circular Lake problem
• The Circular Target Inscribed in a Rectangle dart problem
• Birthday problems: Probability of friends sharing or having different birthdays
• Board exam presentation templates and common traps

1. Defective Batches and Quality Control Sampling

In an industrial batch, items are divided into two mutually exclusive groups: Good Items and Defective Items: $\mathbf{n(S) = \text{Number of Good Items} + \text{Number of Defective Items}}$

    Batch of Items (Total N) = [ Good Items (G) ] + [ Defective Items (D) ]
    Probability of Good Item  = G / N
    Probability of Defective  = D / N = 1 - (G / N)

2. The Non-Replacement Sample Space Trap

When an item is drawn and NOT REPLACED:

• The total sample space decreases by $1$: $n(S') = n(S) - 1$.
• If a good item was removed, the number of remaining good items decreases by $1$.
• If a defective item was removed, the number of remaining defective items decreases by $1$.

3. Geometric Probability (2D Area Models)

In discrete probability (coins, dice, cards), outcomes are countable integers. But what if an outcome can land anywhere across a continuous two-dimensional surface—like a dart thrown at a board, or a rescue helicopter making an emergency landing in a forest?

Definition of Geometric Probability

When the outcomes of an experiment are infinite points distributed uniformly across a geometric region, the probability of an event $E$ occurring within a sub-region is defined as: $\mathbf{P(E) = \frac{\text{Area of the Favourable Sub-Region}}{\text{Total Area of the Entire Sample Space Region}}}$

4. Solved CBSE Board Examination Problems

Solved Example 1: The Non-Replacement Bulb Problem (NCERT Classic)

Problem: (i) A lot of $20$ bulbs contain $4$ defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

Part (i): First Draw

1. Total bulbs: $n(S) = 20$.
2. Defective bulbs: $n(D) = 4$. Good bulbs: $n(G) = 20 - 4 = 16$.
3. Probability of drawing a defective bulb: $P(\text{Defective}) = \frac{4}{20} = \mathbf{\frac{1}{5} = 0.2}$

Part (ii): Second Draw (Non-Replacement)

1. Analyze the Remaining Lot:
   • The bulb drawn in Part (i) was NOT defective (it was good) and was NOT replaced.
   • Total bulbs remaining: $n(S') = 20 - 1 = \mathbf{19}$.
   • Number of good bulbs remaining: $n(G') = 16 - 1 = \mathbf{15}$.
   • Number of defective bulbs remaining: $n(D) = \mathbf{4}$ (unchanged).
2. Probability that the second bulb is not defective (good): $P(\text{Good}) = \frac{n(G')}{n(S')} = \mathbf{\frac{15}{19}}$
3. Therefore:
   • <u>(i) The probability of drawing a defective bulb is $\frac{1}{5}$</u>.
   • <u>(ii) The probability that the second bulb is not defective is $\frac{15}{19}$</u>.

Solved Example 2: The 90 Numbered Discs Problem (CBSE Classic)

Problem: A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears: (i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by $5$

Solution: Total number of discs: $n(S) = 90$ (numbers from $1, 2, 3, \dots, 90$).

1. (i) A two-digit number:

   • Single-digit numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$ ($9$ numbers).
   • Two-digit numbers start from $10$ to $90$: $n(E_1) = 90 - 9 = \mathbf{81}$ $P(\text{Two-digit number}) = \frac{81}{90} = \mathbf{\frac{9}{10} = 0.9}$

2. (ii) A perfect square number:

   • Perfect squares between $1$ and $90$ are: $1^2=1, \; 2^2=4, \; 3^2=9, \; 4^2=16, \; 5^2=25, \; 6^2=36, \; 7^2=49, \; 8^2=64, \; 9^2=81$
   • Total perfect squares: $n(E_2) = \mathbf{9}$. $P(\text{Perfect square}) = \frac{9}{90} = \mathbf{\frac{1}{10} = 0.1}$

3. (iii) A number divisible by 5:

   • Numbers divisible by 5 are: $5, 10, 15, 20, \dots, 90$.
   • Count $= \frac{90}{5} = \mathbf{18}$ numbers. $P(\text{Divisible by 5}) = \frac{18}{90} = \mathbf{\frac{1}{5} = 0.2}$

Solved Example 3: The Helicopter Crash Lake Problem (Geometric Area)

Problem: A missing helicopter is reported to have crashed somewhere in the rectangular region shown in the figure. The region is $9\text{ km}$ long and $4.5\text{ km}$ wide. Inside the region lies a lake of length $2.5\text{ km}$ and width $2\text{ km}$. What is the probability that it crashed inside the lake?

    +------------------------- 9 km --------------------------+
    |                                                         |
    |          +-------- 2.5 km --------+                     | 4.5 km
    |          |          LAKE          | 2 km                |
    |          +------------------------+                     |
    +---------------------------------------------------------+

Solution:

1. Calculate Total Area of the Sample Space (Entire Rectangular Region): $\text{Total Area } A_{\text{total}} = \text{Length} \times \text{Breadth} = 9\text{ km} \times 4.5\text{ km} = \mathbf{40.5\text{ km}^2}$
2. Calculate Area of the Favourable Region (The Lake): $\text{Area of Lake } A_{\text{lake}} = 2.5\text{ km} \times 2\text{ km} = \mathbf{5.0\text{ km}^2}$
3. Apply the Geometric Probability Formula: $P(\text{Crash in Lake}) = \frac{\text{Area of Lake}}{\text{Total Rectangular Area}} = \frac{5.0}{40.5} = \frac{50}{405} = \mathbf{\frac{10}{81}}$
4. Therefore, <u>the probability that the helicopter crashed inside the lake is $\frac{10}{81}$</u>.

5. Summary and Examination Tips

Exam Tip: In the 90 discs problem, remember that $1$ IS a perfect square ($1^2 = 1$). Many students forget to count $1$ and write $8/90$ instead of $9/90$!

Common Mistake: In non-replacement problems, calculating the second probability with the original denominator ($20$). You must reduce the denominator to $19$!