JEE Advanced 1986

LEVELJEE Advanced

Animated Solution for Mathematics - Probability: A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?

Visualized Solution

Defining the Events

Total articles in the lot = $20$
Let $A_{1}$ be the event that the lot has $2$ defective articles: $P (A_{1}) = 0.4$
Let $A_{2}$ be the event that the lot has $3$ defective articles: $P (A_{2}) = 0.6$

The Termination Condition

Let $E$ be the event that testing ends at the $1 2^{t h}$ draw.
Condition for $E$ : The $1 2^{t h}$ article drawn must be the last defective article in the lot.

Case 1: Exactly $2$ Defectives

Case 1: Lot has $2$ defectives ( $A_{1}$ ).
To end at $1 2^{t h}$ draw, $1$ defective must be in the first $11$ draws.
The $2^{n d}$ defective must be exactly at the $1 2^{t h}$ draw.

Probability $P (E ∣ A_{1})$ Setup

Probability of $1$ defective in first $11$ draws: $\frac{^{2} C _{1} \times ^{18} C _{10}}{^{20} C _{11}}$
Probability of drawing the $2^{n d}$ defective on the $1 2^{t h}$ draw: $\frac{1}{9}$
$P (E ∣ A_{1}) = \frac{^{2} C _{1} \times ^{18} C _{10}}{^{20} C _{11}} \times \frac{1}{9}$

Simplifying $P (E ∣ A_{1})$

$P (E ∣ A_{1}) = \frac{2 \times \frac{18 !}{10 ! \times 8 !}}{\frac{20 !}{11 ! \times 9 !}} \times \frac{1}{9}$
$P (E ∣ A_{1}) = \frac{2 \times 11 \times 9}{20 \times 19 \times 9}$
$P (E ∣ A_{1}) = \frac{11}{190}$

Case 2: Exactly $3$ Defectives

Case 2: Lot has $3$ defectives ( $A_{2}$ ).
To end at $1 2^{t h}$ draw, $2$ defectives must be in the first $11$ draws.
The $3^{r d}$ defective must be exactly at the $1 2^{t h}$ draw.

Probability $P (E ∣ A_{2})$ Setup

Probability of $2$ defectives in first $11$ draws: $\frac{^{3} C _{2} \times ^{17} C _{9}}{^{20} C _{11}}$
Probability of drawing the $3^{r d}$ defective on the $1 2^{t h}$ draw: $\frac{1}{9}$
$P (E ∣ A_{2}) = \frac{^{3} C _{2} \times ^{17} C _{9}}{^{20} C _{11}} \times \frac{1}{9}$

Simplifying $P (E ∣ A_{2})$

$P (E ∣ A_{2}) = \frac{3 \times \frac{17 !}{9 ! \times 8 !}}{\frac{20 !}{11 ! \times 9 !}} \times \frac{1}{9}$
$P (E ∣ A_{2}) = \frac{3 \times 11 \times 10 \times 9}{20 \times 19 \times 18 \times 9}$
$P (E ∣ A_{2}) = \frac{11}{228}$

Total Probability Theorem

Using the Law of Total Probability:
$P (E) = P (A_{1}) P (E ∣ A_{1}) + P (A_{2}) P (E ∣ A_{2})$

Substitution of Values

$P (E) = 0.4 \times \frac{11}{190} + 0.6 \times \frac{11}{228}$
$P (E) = \frac{4}{10} \times \frac{11}{190} + \frac{6}{10} \times \frac{11}{228}$

Final Arithmetic

$P (E) = \frac{44}{1900} + \frac{66}{2280}$
$P (E) = \frac{11}{475} + \frac{11}{380}$
$P (E) = \frac{44 + 55}{1900}$

The Final Result

$P (E) = \frac{99}{1900}$
Key Takeaway: The testing ends at the $k^{t h}$ draw if the last defective is found at that exact position.

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The Sigma Insight: Total Probability Theorem

Solution Diagram

Analyzing the Setup

We are given a lot of

20

articles. There are two mutually exclusive hypotheses regarding the number of defective items: $A_{1}$ : The lot contains $2$ defectives, with $P (A_{1}) = 0.4$ .

A_{2}

: The lot contains

3

defectives, with

P (A_{2}) = 0.6

The testing procedure terminates exactly at the

1 2^{t h}

draw. This implies that the final defective item must be drawn on the

1 2^{t h}

attempt.

Case 1

The Lot has 2 Defectives

For the process to end at the

1 2^{t h}

draw, we must have found exactly

1

defective in the first

11

draws, and the

2^{n d}

defective must appear on the

1 2^{t h}

draw.

The probability of finding

1

defective in the first

11

draws is given by the hypergeometric distribution:

P (1 def in 11) = \frac{^{2} C _{1} \times ^{18} C _{10}}{^{20} C _{11}}

After

11

draws,

9

articles remain, one of which is the final defective. The probability of picking it is

\frac{1}{9}

. Thus, the conditional probability is:

P (E ∣ A_{1}) = (\frac{^{2} C _{1} \times ^{18} C _{10}}{^{20} C _{11}}) \times \frac{1}{9} = \frac{11}{190}

Case 2

The Lot has 3 Defectives

For the process to end at the

1 2^{t h}

draw, we must have found exactly

2

defectives in the first

11

draws, and the

3^{r d}

defective must appear on the

1 2^{t h}

draw.

The probability of finding

2

defectives in the first

11

draws is:

P (2 def in 11) = \frac{^{3} C _{2} \times ^{17} C _{9}}{^{20} C _{11}}

Multiplying by the probability of picking the final defective from the remaining

9

items, we get:

P (E ∣ A_{2}) = (\frac{^{3} C _{2} \times ^{17} C _{9}}{^{20} C _{11}}) \times \frac{1}{9} = \frac{11}{228}

Final Calculation

We invoke the Law of Total Probability to combine these cases:

P (E) = P (A_{1}) P (E ∣ A_{1}) + P (A_{2}) P (E ∣ A_{2})

Substituting the known values:

P (E) = 0.4 \times (\frac{11}{190}) + 0.6 \times (\frac{11}{228})

Performing the arithmetic, we arrive at the final result: $P (E) = \frac{99}{1900}$

Common Traps & FAQs

Wait, why is the probability of the last draw 1/9? Shouldn't it be more complex?

Great question! After 11 draws, we have 20 - 11 = 9 articles remaining in the lot. If we have successfully found all but one defective in the first 11 draws, that final defective is sitting somewhere among the remaining 9 articles. Since we are drawing randomly, the probability that the specific remaining defective is the one we pick next is simply 1 out of 9.

Why can't we use Binomial Distribution here?

Binomial distribution requires independent trials with constant probability (sampling with replacement). Here, we are sampling without replacement. The probability of picking a defective changes with every draw because the composition of the lot changes. Combinations are your best friend here.

Why do we split the problem into two cases?

Because the number of defective articles is not fixed; it's a random variable with two possible states. We don't know if we are dealing with a lot of 2 or 3 defectives. The Law of Total Probability allows us to weigh the outcome of each scenario by the likelihood of that scenario occurring.

Visualized Solution

Defining the Events

The Termination Condition

Case 1: Exactly $2$ Defectives

Probability $P (E ∣ A_{1})$ Setup

Simplifying $P (E ∣ A_{1})$

Case 2: Exactly $3$ Defectives

Probability $P (E ∣ A_{2})$ Setup

Simplifying $P (E ∣ A_{2})$

Total Probability Theorem

Substitution of Values

Final Arithmetic

The Final Result

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

Case 1

Case 2

Final Calculation

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Visualized Solution

Defining the Events

The Termination Condition

Case 1: Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 Defectives

Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣A1​) Setup

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣A1​)

Case 2: Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 Defectives

Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣A2​) Setup

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣A2​)

Total Probability Theorem

Substitution of Values

Final Arithmetic

The Final Result

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

Case 1

Case 2

Final Calculation

Similar Questions

A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)

A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

Case 1: Exactly $2$ Defectives

Probability $P (E ∣ A_{1})$ Setup

Simplifying $P (E ∣ A_{1})$

Case 2: Exactly $3$ Defectives

Probability $P (E ∣ A_{2})$ Setup

Simplifying $P (E ∣ A_{2})$