JEE Main 2022 (25 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Select Answer:

Visualized Solution

Parameters of Binomial Distribution

Let the mean be $μ = n p$
Let the variance be $σ^{2} = n pq$

Given Conditions

Sum of mean and variance: $n p + n pq = 24$
Product of mean and variance: $n p \cdot n pq = 128$

Forming the Quadratic Equation

If sum $S$ and product $P$ of two numbers are known, they are roots of $t^{2} - S t + P = 0$
Substituting $S = 24$ and $P = 128$ :
Equation: $t^{2} - 24 t + 128 = 0$

Solving for Roots

Factorizing: $t^{2} - 16 t - 8 t + 128 = 0$
$(t - 16) (t - 8) = 0$
Roots: $t = 16, 8$

Identifying Mean and Variance

In Binomial Distribution, probability of failure $0 \leq q \leq 1$
Therefore, variance is always less than or equal to mean: $n pq \leq n p$
Mean $n p = 16$
Variance $n pq = 8$

Finding Probabilities $p$ and $q$

$q = \frac{n pq}{n p} = \frac{8}{16} = \frac{1}{2}$
Since $p + q = 1$ , $p = 1 - \frac{1}{2} = \frac{1}{2}$

Calculating Number of Trials $n$

We know Mean $n p = 16$
Substitute $p = \frac{1}{2}$ : $n \cdot \frac{1}{2} = 16$
Total trials $n = 32$

Probability of One or Two Successes

Required Probability: $P (X = 1) + P (X = 2)$
Binomial Formula: $P (X = r) = (r n) p^{r} q^{n - r}$

Calculating Individual Probabilities

$P (X = 1) = (1 32) (\frac{1}{2})^{1} (\frac{1}{2})^{31} = 32 \cdot \frac{1}{2 ^{32}}$
$P (X = 2) = (2 32) (\frac{1}{2})^{2} (\frac{1}{2})^{30} = \frac{32 \times 31}{2} \cdot \frac{1}{2 ^{32}} = 496 \cdot \frac{1}{2 ^{32}}$

Final Result Calculation

Sum $= \frac{32 + 496}{2 ^{32}} = \frac{528}{2 ^{32}}$
Factorizing 528: $528 = 33 \times 16 = 33 \times 2^{4}$
Probability $= \frac{33 \times 2 ^{4}}{2 ^{32}} = \frac{33}{2 ^{28}}$

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

In a binomial distribution, the mean is defined as

μ = n p

and the variance is defined as

σ^{2} = n pq

. We are given the sum and the product of these two parameters:

n p + n pq = 24

n p \cdot n pq = 128

The Master Equation

When given the sum

S

and product

P

of two quantities, they serve as the roots of the quadratic equation

t^{2} - S t + P = 0

. Substituting our values, we obtain:

t^{2} - 24 t + 128 = 0

Factoring this quadratic, we look for two numbers that multiply to

128

and add to

24

. These numbers are

16

and

8

. Thus, the roots are

t = 16

and

t = 8

Applying Physical Constraints

We must determine which value corresponds to the mean and which to the variance. Recall that the variance is

σ^{2} = n pq

, where

q

is the probability of failure (

0 < q < 1

Because

q < 1

, it follows that

n pq < n p

. Therefore, the mean must be the larger value, and the variance must be the smaller value:

n p = 16

n pq = 8

Solving for Parameters

We find

q

by taking the ratio of the variance to the mean:

q = \frac{n pq}{n p} = \frac{8}{16} = \frac{1}{2}

Since

p + q = 1

, we find

p = 1 - \frac{1}{2} = \frac{1}{2}

. Substituting

p = \frac{1}{2}

into the mean equation

n p = 16

n \cdot \frac{1}{2} = 16 \Rightarrow n = 32

Final Calculation

We calculate the probability of one or two successes using the binomial formula

P (X = r) = (r n) p^{r} q^{n - r}

For

r = 1

P (X = 1) = (1 32) (\frac{1}{2})^{1} (\frac{1}{2})^{31} = 32 \cdot \frac{1}{2 ^{32}}

For

r = 2

P (X = 2) = (2 32) (\frac{1}{2})^{2} (\frac{1}{2})^{30} = \frac{32 \times 31}{2} \cdot \frac{1}{2 ^{32}} = 496 \cdot \frac{1}{2 ^{32}}

Summing these probabilities:

P (X = 1) + P (X = 2) = \frac{32 + 496}{2 ^{32}} = \frac{528}{2 ^{32}}

Factoring

528

33 \times 16

(or

33 \times 2^{4}

), we simplify the expression to reach the final answer:

\frac{33 \times 2 ^{4}}{2 ^{32}} = \frac{33}{2 ^{28}}

Common Traps & FAQs

Why do we assume the variance is smaller than the mean? Is it possible for variance to be larger?

This is a fantastic conceptual question. In a binomial distribution, the variance is defined as

σ^{2} = n pq

. Since

q

represents the probability of failure, it must lie in the interval

[0, 1]

. Therefore, multiplying the mean

n p

by a factor

q

(which is

\leq 1

) will always result in a value less than or equal to the mean. It is mathematically impossible for the variance to exceed the mean in a standard binomial distribution.

Is there a faster way to solve the quadratic equation

t^{2} - 24 t + 128 = 0

While the quadratic formula is always a reliable fallback, looking for factors is usually faster in JEE exams. You need two numbers that multiply to 128 and add to 24. Think of powers of 2: 128 is

2^{7}

. You can test pairs like

8 \times 16

. Since

8 + 16 = 24

, you have your roots instantly. Developing this 'number sense' saves precious seconds during the exam.

Why did we add the probabilities for

X = 1

and

X = 2

The question asks for the probability of 'one or two successes'. In probability theory, the word 'or' signifies the union of events. Since getting exactly one success and getting exactly two successes are mutually exclusive events (you cannot have both simultaneously in a single trial set), the probability of the union is simply the sum of the individual probabilities:

P (X = 1 \cup X = 2) = P (X = 1) + P (X = 2)

Animated Solution for Mathematics - Probability: If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Select Answer:

Visualized Solution

Parameters of Binomial Distribution

Given Conditions

Forming the Quadratic Equation

Solving for Roots

Identifying Mean and Variance

Finding Probabilities $p$ and $q$

Calculating Number of Trials $n$

Probability of One or Two Successes

Calculating Individual Probabilities

Final Result Calculation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Master Equation

Applying Physical Constraints

Solving for Parameters

Final Calculation

Similar Questions

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is:

Consider 5 independent Bernoulli's trials each with probability of success $p$ . If the probability of at least one failure is greater than or equal to $31/32$ , then $p$ lies in the interval

A random variable $X$ has the following probability distribution: X: 1, 2, 3, 4, 5; P(X): $K^{2}, 2 K, K, 2 K, 5 K^{2}$ . Then $P (X > 2)$ is equal to:

Let the mean and the standard deviation of the probability distribution $X P (X) α \frac{1}{3} 1 K 0 \frac{1}{6} - 3 \frac{1}{4}$ be $μ$ and $σ$ , respectively. If $σ - μ = 2$ , then $σ + μ$ is equal to________

Let $X$ be a random variable with distribution. $x P (X = x) - 2 \frac{1}{5} - 1 a 3 \frac{1}{3} 4 \frac{1}{5} 6 b$ If the mean of $X$ is 2.3 and variance of $X$ is $σ^{2}$ , then $100 σ^{2}$ is equal to :

Let $X$ be a random variable such that the probability function of a distribution is given by $P (X = 0) = 1/2, P (X = j) = 1/ 3^{j} (j = 1, 2, 3, \dots, \infty)$ . Then the mean of the distribution and $P (X is positive and even)$ respectively are:

A random variable $X$ has the following probability distribution: $X : 1, 2, 3, 4, 5$ $P (x) : K^{2}, 2 k, K, 2 K, 5 K^{2}$ Then $P (x > 2)$ is equal to :

If the mean of the following probability distribution of a random variable $X$ : $X P (X) 0 a 2 2 a 4 a + b 6 2 b 8 3 b$ is $46/9$ , then the variance of the distribution is

If the probability that the random variable $X$ takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to

Animated Solution for Mathematics - Probability: If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Select Answer:

Visualized Solution

Parameters of Binomial Distribution

Given Conditions

Forming the Quadratic Equation

Solving for Roots

Identifying Mean and Variance

Finding Probabilities .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Calculating Number of Trials .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Probability of One or Two Successes

Calculating Individual Probabilities

Final Result Calculation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Master Equation

Applying Physical Constraints

Solving for Parameters

Final Calculation

Similar Questions

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A random variable X has the following probability distribution: X: 1, 2, 3, 4, 5; P(X): K2,2K,K,2K,5K2. Then P(X>2) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A random variable X has the following probability distribution: X:1,2,3,4,5 P(x):K2,2k,K,2K,5K2 Then P(x>2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the following probability distribution of a random variable X: XP(X)​0a​22a​4a+b​62b​83b​​ is 46/9, then the variance of the distribution is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the probability that the random variable X takes values x is given by P(X=x)=k(x+1)3−x,x=0,1,2,3,…, where k is a constant, then P(X≥2) is equal to

Finding Probabilities $p$ and $q$

Calculating Number of Trials $n$

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is:

A random variable $X$ has the following probability distribution: X: 1, 2, 3, 4, 5; P(X): $K^{2}, 2 K, K, 2 K, 5 K^{2}$ . Then $P (X > 2)$ is equal to:

A random variable $X$ has the following probability distribution: $X : 1, 2, 3, 4, 5$ $P (x) : K^{2}, 2 k, K, 2 K, 5 K^{2}$ Then $P (x > 2)$ is equal to :

If the mean of the following probability distribution of a random variable $X$ : $X P (X) 0 a 2 2 a 4 a + b 6 2 b 8 3 b$ is $46/9$ , then the variance of the distribution is

If the probability that the random variable $X$ takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to