JEE Main 2023 (11 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let the probability of getting head for a biased coin be $\frac{1}{4}$ . It is tossed repeatedly until a head appears. Let $N$ be the number of tosses required. If the probability that the equation $64 x^{2} + 5 N x + 1 = 0$ has no real root is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q - p$ is equal to

Q: Wait, why do we need the discriminant to be strictly less than zero? Can't it be equal to zero?

Great question! If the discriminant $D$ is equal to zero, the quadratic equation has exactly one real root (a repeated root). Graphically, this means the parabola touches the x-axis at exactly one point. Since the problem asks for the condition where there are *no* real roots, we must ensure the parabola never touches or crosses the x-axis, which only happens when $D < 0$.

Q: Why is $N$ restricted to integers? Can't $N$ be a continuous variable?

In this experiment, $N$ represents the number of tosses required to get the first head. Since you cannot perform a fraction of a coin toss, $N$ must be a discrete, positive integer. We are counting events, not measuring continuous time or distance, so $N$ must belong to the set $\{1, 2, 3, \dots\}$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): To determine if the quadratic equation

64 x^{2} + 5 N x + 1 = 0

has no real roots, you must focus on the discriminant

D = b^{2} - 4 a c

. Visualize the parabola

y = 64 x^{2} + 5 N x + 1

; for it to have no real roots, it must never touch the x-axis, meaning it must lie entirely above it.

Hint 2

Hint 2 (The Arsenal): Recall that for the quadratic equation

a x^{2} + b x + c = 0

, the condition for no real roots is

D < 0

. Substitute your coefficients

a = 64

b = 5 N

, and

c = 1

into this inequality. Remember that

N

represents the number of tosses, so

N

must be a positive integer (

N \in {1, 2, 3, \dots}

Hint 3

Hint 3 (The Execution): After setting up the inequality

(5 N)^{2} - 4 (64) (1) < 0

, you will arrive at

25 N^{2} < 256

. Solve for

N^{2}

and identify all positive integers

N

that satisfy this condition. Once you have the valid values for

N

, calculate the probability of each scenario using the geometric distribution formula

P (N = k) = P (T)^{k - 1} \cdot P (H)

Visualized Solution

Understanding the Experiment

Biased coin probabilities: $P (H) = \frac{1}{4}$ , $P (T) = \frac{3}{4}$
Experiment: Toss repeatedly until a Head ( $H$ ) appears.
Let $N$ be the total number of tosses required.
Target Equation: $64 x^{2} + 5 N x + 1 = 0$

Condition for No Real Roots

For a quadratic equation $a x^{2} + b x + c = 0$ to have no real roots, its discriminant must be negative.
Discriminant formula: $D = b^{2} - 4 a c < 0$
Graphically, the parabola must not intersect the x-axis.

Setting up the Discriminant

Compare with $a x^{2} + b x + c = 0$ :
$a = 64$
$b = 5 N$
$c = 1$
Substitute into $D < 0$ : $(5 N)^{2} - 4 (64) (1) < 0$

Solving the Inequality

Expand the terms: $25 N^{2} - 256 < 0$
Rearrange: $25 N^{2} < 256$
Isolate $N^{2}$ : $N^{2} < \frac{256}{25}$
Calculate the decimal value: $N^{2} < 10.24$

Finding Valid Toss Counts

$N$ represents the number of coin tosses, so $N \in {1, 2, 3, \dots}$
Check $N = 1$ : $1^{2} = 1 < 10.24$ (Valid)
Check $N = 2$ : $2^{2} = 4 < 10.24$ (Valid)
Check $N = 3$ : $3^{2} = 9 < 10.24$ (Valid)
Check $N = 4$ : $4^{2} = 16 > 10.24$ (Invalid)
Therefore, $N$ can only be $1, 2,$ or $3$ .

Probability of $N$ Tosses

The event $N = k$ means getting $(k - 1)$ Tails followed by $1$ Head.
Since tosses are independent, we multiply their probabilities.
General formula: $P (N = k) = P (T)^{k - 1} \cdot P (H)$
$P (N = k) = (\frac{3}{4})^{k - 1} \cdot (\frac{1}{4})$

Probability for $N = 1$

$N = 1$ means getting a Head on the very first toss.
$P (N = 1) = P (H)$
$P (N = 1) = \frac{1}{4}$

Probability for $N = 2$

$N = 2$ means getting a Tail first, then a Head.
$P (N = 2) = P (T) \cdot P (H)$
$P (N = 2) = (\frac{3}{4}) \cdot (\frac{1}{4}) = \frac{3}{16}$

Probability for $N = 3$

$N = 3$ means getting two Tails, followed by a Head.
$P (N = 3) = P (T)^{2} \cdot P (H)$
$P (N = 3) = (\frac{3}{4})^{2} \cdot (\frac{1}{4})$
$P (N = 3) = (\frac{9}{16}) \cdot (\frac{1}{4}) = \frac{9}{64}$

Total Required Probability

The events $N = 1, N = 2,$ and $N = 3$ are mutually exclusive.
Total Probability $P = P (N = 1) + P (N = 2) + P (N = 3)$
$P = \frac{1}{4} + \frac{3}{16} + \frac{9}{64}$
Make denominators equal: $P = \frac{16}{64} + \frac{12}{64} + \frac{9}{64}$
$P = \frac{37}{64}$

Finding $q - p$

We found the probability is $\frac{37}{64}$
The problem states this probability is $\frac{p}{q}$
Comparing them: $p = 37$ and $q = 64$
Check: $37$ and $64$ are co-prime (no common factors).
Calculate $q - p = 64 - 37 = 27$
Final Answer: 27

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

We are given a biased coin where the probability of heads is

P (H) = \frac{1}{4}

and the probability of tails is

P (T) = \frac{3}{4}

. The experiment involves tossing the coin until the first head appears, with

N

representing the total number of tosses.

We are tasked with finding the probability that the quadratic equation

64 x^{2} + 5 N x + 1 = 0

has no real roots.

The Algebraic Gatekeeper

A quadratic equation

a x^{2} + b x + c = 0

has no real roots if and only if its discriminant

D = b^{2} - 4 a c

is strictly less than zero. For our equation, the coefficients are

a = 64

b = 5 N

, and

c = 1

Substituting these into the discriminant condition:

D = (5 N)^{2} - 4 (64) (1) < 0

25 N^{2} - 256 < 0

25 N^{2} < 256

N^{2} < \frac{256}{25} = 10.24

The Discrete Reality

Since

N

represents the number of tosses, it must be a positive integer (

N \in {1, 2, 3, ...}

). We test the possible values of

N

that satisfy

N^{2} < 10.24

For

N = 1

1^{2} = 1 < 10.24

(Valid)

For

N = 2

2^{2} = 4 < 10.24

(Valid)

For

N = 3

3^{2} = 9 < 10.24

(Valid)

For

N = 4

4^{2} = 16 > 10.24

(Invalid)

Thus, the condition for no real roots is satisfied if

N \in {1, 2, 3}

The Probability Tree

The probability that the first head appears on the

k

-th toss follows the geometric distribution:

P (N = k) = P (T)^{k - 1} \cdot P (H)

. We calculate the probabilities for our valid cases:

For

N = 1

P (N = 1) = \frac{1}{4}

For

N = 2

P (N = 2) = (\frac{3}{4}) \cdot (\frac{1}{4}) = \frac{3}{16}

For

N = 3

P (N = 3) = (\frac{3}{4})^{2} \cdot (\frac{1}{4}) = \frac{9}{16} \cdot \frac{1}{4} = \frac{9}{64}

Final Calculation

Since these events are mutually exclusive, the total probability is the sum of the individual probabilities:

P = \frac{1}{4} + \frac{3}{16} + \frac{9}{64}

Converting to a common denominator of

64

P = \frac{16}{64} + \frac{12}{64} + \frac{9}{64} = \frac{37}{64}

Given

P = \frac{p}{q}

where

p = 37

and

q = 64

are co-prime, we find the final result:

q - p = 64 - 37 = 27

Common Traps & FAQs

Wait, why do we need the discriminant to be strictly less than zero? Can't it be equal to zero?

Great question! If the discriminant

D

is equal to zero, the quadratic equation has exactly one real root (a repeated root). Graphically, this means the parabola touches the x-axis at exactly one point. Since the problem asks for the condition where there are *no* real roots, we must ensure the parabola never touches or crosses the x-axis, which only happens when

D < 0

Why is

N

restricted to integers? Can't

N

be a continuous variable?

In this experiment,

N

represents the number of tosses required to get the first head. Since you cannot perform a fraction of a coin toss,

N

must be a discrete, positive integer. We are counting events, not measuring continuous time or distance, so

N

must belong to the set

{1, 2, 3, \dots}

How do I know when to add the probabilities of

N = 1, N = 2,

and

N = 3

We add them because these events are mutually exclusive. You cannot have the first head appear on the first toss *and* on the second toss simultaneously. In probability, when you are looking for the chance of one event *or* another event occurring, and those events cannot happen at the same time, you sum their individual probabilities.

Enter Numerical Value:

Visualized Solution

Understanding the Experiment

Condition for No Real Roots

Setting up the Discriminant

Solving the Inequality

Finding Valid Toss Counts

Probability of $N$ Tosses

Probability for $N = 1$

Probability for $N = 2$

Probability for $N = 3$

Total Required Probability

Finding $q - p$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Algebraic Gatekeeper

The Discrete Reality

The Probability Tree

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Understanding the Experiment

Condition for No Real Roots

Setting up the Discriminant

Solving the Inequality

Finding Valid Toss Counts

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

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Total Required Probability

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The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Algebraic Gatekeeper

The Discrete Reality

The Probability Tree

Final Calculation

Similar Questions

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is .........

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A fair n(n>1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is 9n​, then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is:

Probability of $N$ Tosses

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Finding $q - p$

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