JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $1/2$ and probability of occurrence of 0 at the odd place be $1/3$ . Then the probability that '10' is followed by '01' is equal to :

Q: Why is the probability of 1 at an odd place 2/3?

The problem states the probability of 0 at an odd place is $1/3$. Since the only other option is 1, the sum of probabilities must be 1. Therefore, $P(1_O) = 1 - P(0_O) = 1 - 1/3 = 2/3$. Always remember: the sum of all possible outcomes in a single trial is always 1.

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Visualized Solution

Decoding the Target Sequence

The problem asks for the probability that `'10'` is followed by `'01'`.
This means we are looking for the exact $4$ -digit sequence: `'1001'`.
The probabilities of getting $0$ or $1$ depend on whether the position is odd or even.

Analyzing the Given Probabilities

For Even places: $P (0_{E}) = \frac{1}{2} ⟹ P (1_{E}) = 1 - \frac{1}{2} = \frac{1}{2}$
For Odd places: $P (0_{O}) = \frac{1}{3} ⟹ P (1_{O}) = 1 - \frac{1}{3} = \frac{2}{3}$

Case $1$ - Starting at an Odd Position

The sequence `'1001'` can start at an Odd position.
The positions will be: Odd, Even, Odd, Even.
Digits required: $1$ (Odd), $0$ (Even), $0$ (Odd), $1$ (Even).

Substituting Probabilities for Case $1$

$P (Case 1) = P (1_{O}) \times P (0_{E}) \times P (0_{O}) \times P (1_{E})$
Substitute the values: $\frac{2}{3} \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2}$

Calculating Case $1$ Probability

Multiply the numerators: $2 \times 1 \times 1 \times 1 = 2$
Multiply the denominators: $3 \times 2 \times 3 \times 2 = 36$
$P (Case 1) = \frac{2}{36} = \frac{1}{18}$

Case $2$ - Starting at an Even Position

The sequence `'1001'` could also start at an Even position.
The positions will be: Even, Odd, Even, Odd.
Digits required: $1$ (Even), $0$ (Odd), $0$ (Even), $1$ (Odd).

Substituting Probabilities for Case $2$

$P (Case 2) = P (1_{E}) \times P (0_{O}) \times P (0_{E}) \times P (1_{O})$
Substitute the values: $\frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} \times \frac{2}{3}$

Calculating Case $2$ Probability

Multiply the numerators: $1 \times 1 \times 1 \times 2 = 2$
Multiply the denominators: $2 \times 3 \times 2 \times 3 = 36$
$P (Case 2) = \frac{2}{36} = \frac{1}{18}$

Total Probability

The sequence can start at an odd position OR an even position.
Total Probability = $P (Case 1) + P (Case 2)$
Total Probability = $\frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9}$

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Rhythm

Before we dive into the math, let's look at the rules of our machine. We are given the probability of a

0

appearing at even positions as

P (0_{E}) = \frac{1}{2}

, which implies

P (1_{E}) = 1 - \frac{1}{2} = \frac{1}{2}

At odd positions, the machine is more restrictive:

P (0_{O}) = \frac{1}{3}

, which means

P (1_{O}) = 1 - \frac{1}{3} = \frac{2}{3}

Think of these as the 'DNA' of our sequence. Every time we place a digit, we must check the index. This dependency on the index is the core of the problem.

The Two Paths of Destiny

When we look for the sequence '1001', we must realize that the probability of the sequence occurring depends entirely on whether it starts at an odd index or an even index.

If the sequence starts at an odd index, the positions follow the pattern: Odd, Even, Odd, Even.

If the sequence starts at an even index, the pattern shifts: Even, Odd, Even, Odd. These are our two mutually exclusive cases.

Calculating the Probabilities

Let's calculate the probability for Case 1, where the sequence '1001' starts at an odd position:

P (Case 1) = P (1_{O}) \times P (0_{E}) \times P (0_{O}) \times P (1_{E})

Substituting our values, we get:

P (Case 1) = \frac{2}{3} \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{2}{36} = \frac{1}{18}

Now, let's evaluate Case 2, where the sequence '1001' starts at an even position:

P (Case 2) = P (1_{E}) \times P (0_{O}) \times P (0_{E}) \times P (1_{O})

Substituting these values:

P (Case 2) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} \times \frac{2}{3} = \frac{2}{36} = \frac{1}{18}

The Grand Synthesis

Since the sequence can start at an odd position OR an even position, we must add these probabilities together to find the total probability of the event occurring.

Total Probability = P (Case 1) + P (Case 2)

Total Probability = \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9}

The final probability of the sequence '1001' appearing is $\frac{1}{9}$ . You have successfully mapped the behavior of the system and decoded the rhythm of the binary string.

Common Traps & FAQs

Wait, why do we have to consider two cases? Can't we just multiply the probabilities directly?

Great question! If the probability of getting a 0 were the same for every position, you could just multiply. But here, the 'rules of the game' change based on the position index. Since the sequence '1001' can begin at either an odd or an even index, these are two distinct scenarios that both satisfy the condition. We must calculate them separately and add them because they are mutually exclusive.

How do I know if the position is odd or even? Does the problem tell me?

The problem defines the probability based on the position index. In probability problems like this, we assume the sequence is indexed

1, 2, 3, 4, \dots

. If the sequence starts at index

n

, then

n

is either odd or even. You don't need to know the specific value of

n

, just its parity!

Why is the probability of 1 at an odd place 2/3?

The problem states the probability of 0 at an odd place is

1/3

. Since the only other option is 1, the sum of probabilities must be 1. Therefore,

P (1_{O}) = 1 - P (0_{O}) = 1 - 1/3 = 2/3

. Always remember: the sum of all possible outcomes in a single trial is always 1.

Is there a trick to the '10 followed by 01' phrasing?

It's a classic JEE phrasing designed to make you overthink! '10 followed by 01' is just a fancy way of saying the 4-digit string '1001'. Whenever you see 'followed by', just concatenate the strings and focus on the resulting pattern.

What if the sequence was longer, like '10010'? Would the logic change?

The logic remains identical! You would still identify the parity of the starting position and then map out the parity of every subsequent position. The complexity would increase, but the fundamental 'Case 1 vs Case 2' approach is your most powerful tool.

Select Answer:

Visualized Solution

Decoding the Target Sequence

Analyzing the Given Probabilities

Case $1$ - Starting at an Odd Position

Substituting Probabilities for Case $1$

Calculating Case $1$ Probability

Case $2$ - Starting at an Even Position

Substituting Probabilities for Case $2$

Calculating Case $2$ Probability

Total Probability

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Rhythm

The Two Paths of Destiny

Calculating the Probabilities

The Grand Synthesis

Similar Questions

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Select Answer:

Visualized Solution

Decoding the Target Sequence

Analyzing the Given Probabilities

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 - Starting at an Odd Position

Substituting Probabilities for Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

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

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 - Starting at an Even Position

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

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

Total Probability

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Rhythm

The Two Paths of Destiny

Calculating the Probabilities

The Grand Synthesis

Similar Questions

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If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is

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Case $1$ - Starting at an Odd Position

Substituting Probabilities for Case $1$

Calculating Case $1$ Probability

Case $2$ - Starting at an Even Position

Substituting Probabilities for Case $2$

Calculating Case $2$ Probability

The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is