A six faced die is biased such that 3 × P(a prime number) = 6 × P(a composite number) = 2 × P(1). Le

Q: Wait, why is 1 not a prime number? I thought it was prime.

Ah, the classic trap! Remember, a prime number must have exactly two distinct factors: 1 and itself. Since 1 only has one factor, it is neither prime nor composite. It is 'unitary'. Always be careful with 1 in number theory problems!

Q: Why did we set the relation to $6k$ instead of just $k$?

Great question. We could have used $k$, but that would force us to work with fractions like $k/3$ and $k/6$ immediately. By choosing the LCM of the coefficients (3, 6, 2), which is 6, we ensure that every probability is an integer multiple of $k$. It is a small optimization that prevents big mistakes.

Q: Can I just calculate the expectation manually using the table instead of using the Binomial formula?

You absolutely can! You could calculate $P(X=0)$, $P(X=1)$, and $P(X=2)$ and sum $x \cdot P(x)$. But why take the long road when the Binomial Distribution gives you the shortcut $E(X) = np$? JEE is about speed and accuracy—embrace the shortcut!

Question

A six faced die is biased such that $3 	imes P(	ext{a prime number}) = 6 	imes P(	ext{a composite number}) = 2 	imes P(1)$. Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is :

Options:

(A) $\frac{3}{11}$

(B) $\frac{5}{11}$

(C) $\frac{7}{11}$

(D) $\frac{8}{11}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we first categorize the outcomes of the die into three distinct groups:
*   The 'Unitary' number: $\{1\}$
*   The 'Prime' numbers: $\{2, 3, 5\}$
*   The 'Composite' numbers: $\{4, 6\}$

This categorization allows us to handle the bias systematically rather than treating each face individually.

### The Algebra of the Bias

The problem provides the following symmetric relation:
$$3 	imes P(	ext{prime}) = 6 	imes P(	ext{composite}) = 2 	imes P(1)$$

To simplify this, we equate the entire expression to a constant $6k$, where $6$ is the Least Common Multiple of $3, 6,$ and $2$. This choice eliminates fractions and allows us to express each probability in terms of $k$:

*   $P(	ext{prime}) = \frac{6k}{3} = 2k$
*   $P(	ext{composite}) = \frac{6k}{6} = k$
*   $P(1) = \frac{6k}{2} = 3k$

### The Law of Total Probability

We invoke the fundamental axiom that the sum of all probabilities must equal $1$. Given that there is one '1', three 'primes', and two 'composites', we write:

$$P(1) + [P(2) + P(3) + P(5)] + [P(4) + P(6)] = 1$$

Substituting our $k$-values into the equation:

$$3k + (2k + 2k + 2k) + (k + k) = 1$$
$$3k + 6k + 2k = 11k = 1$$

Solving for $k$, we find:
$$k = \frac{1}{11}$$

### The Perfect Square Event

The variable $X$ counts the number of perfect squares obtained. In the set $\{1, 2, 3, 4, 5, 6\}$, the perfect squares are $\{1, 4\}$.

The probability of success $p$ in a single throw is:
$$p = P(1) + P(4) = 3k + k = 4k$$

Substituting the value of $k$:
$$p = 4 	imes \frac{1}{11} = \frac{4}{11}$$

### The Binomial Leap

We are performing $n=2$ independent trials. This follows a Binomial Distribution with parameters $n=2$ and $p = \frac{4}{11}$.

The mean (expected value) of a Binomial Distribution is given by $E(X) = np$. Therefore:

$$E(X) = 2 	imes \frac{4}{11} = \frac{8}{11}$$

The final expected value is **$\frac{8}{11}$**.

Select Answer:

Visualized Solution

Sample Space of the Die

Categorizing the Outcomes

Given Probability Relation

Proportionality Constant $k$

Probabilities in terms of $k$

Law of Total Probability

Substituting Probabilities

Solving for $k$

Identifying Perfect Squares

Probability of a Perfect Square

Evaluating $p$

Binomial Distribution Setup

Mean of $X$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Algebra of the Bias

The Law of Total Probability

The Perfect Square Event

The Binomial Leap

Similar Questions

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is

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:

Comprehension Passage

The probability that $X = 3$ equals

The probability that $X \geq 3$ equals

The conditional probability that $X \geq 6$ given $X > 3$ equals

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6∣ X > 3)$ . Then $\frac{b + c}{a}$ is equal to

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 $\frac{n}{9}$ , then $n$ is equal to

A fair die is tossed until six is obtained on it. Let $X$ be the number of required tosses, then the conditional probability $P (X \geq 5∣ X > 2)$ is :

An unbiased coin is tossed 5 times. Suppose that a variable $X$ is assigned the value $k$ when $k$ consecutive heads are obtained for $k = 3, 4, 5$ , otherwise $X$ takes the value $- 1$ . The expected value of $X$ , is

Two dice $A$ and $B$ are rolled. Let the numbers obtained on $A$ and $B$ be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to