Q.(a) A die with numbers 1 to 6 is biased such that $P(2) = \dfrac{3}{10}$ and the probability of other numbers is equal. Find the mean of the number of times the number 2 appears on the die, if the die is thrown twice.

The key is to recall the basic parameters of a binomial distribution $B(n, p)$: the mean is $np$ and the variance is $npq$, where $q = 1-p$. The problem gives two relations between these quantities — their difference and their product — which is enough to pin down $n$ and $p$ uniquely.

Let’s denote mean $M = np$ and variance $V = npq$. We are told:

Since $V = npq = np(1-p)$, the difference $M - V = np - np(1-p) = np^2$. That’s a clean simplification: the difference of mean and variance is just $np^2$. So we have:

And the product is:

Now we have two equations in $n$, $p$, and $q$ (with $q = 1-p$). Let’s work through them step by step.

1. Find $q$ from the ratio. Divide equation (2) by equation (1):

So $n q = 5$. Since $q = 1-p$, this is $n(1-p) = 5$.

2. Express $n$ in terms of $p$.

   From $n q = 5$, we have $n = \frac{5}{q} = \frac{5}{1-p}$.

3. Substitute into equation (1).

   Equation (1) says $n p^2 = \frac{4}{7}$. Replace $n$:

Multiply both sides by $1-p$:

Multiply through by 7:

Bring all terms to one side:

4. Solve the quadratic for $p$. Using the quadratic formula:

The positive root is $p = \frac{20}{70} = \frac{2}{7}$. (The other root is negative, which is invalid for a probability.)

So $p = \frac{2}{7}$, and therefore $q = 1 - \frac{2}{7} = \frac{5}{7}$.

5. Find $n$. …

The key idea here is that a binomial distribution is completely determined by two parameters: the number of trials $n$ and the probability of success $p$ on each trial. The mean and variance are not independent — they are linked by the relation $\text{variance} = \text{mean} \times (1-p)$. Once you extract $p$ from that, $n$ follows directly.

Let’s walk through it.

1. Recall the standard formulas. For a binomial distribution with parameters $n$ and $p$,

where $q = 1-p$ is the probability of failure.

2. Use the given values to find $p$. We are told:

Divide variance by mean:

So $q = \frac{1}{3}$, and therefore

3. Now find $n$. From $np = \frac{10}{3}$ and $p = \frac{2}{3}$,

So we have a binomial distribution with $n=5$ trials and success probability $p = \frac{2}{3}$.

4. Interpret "at least one success". …

For a binomial distribution, mean $=np$ and variance $=npq$ with $q=1-p$. Hence

and a valid distribution needs $0<q<1$ and $n$ a positive integer.

• (A) Mean $6$, var $2$: $q=\tfrac{2}{6}=\tfrac13,\ p=\tfrac23,\ n=\dfrac{6}{2/3}=9$ ✓
• (B) Mean $16$, var $12$: $q=\tfrac{12}{16}=\tfrac34,\ p=\tfrac14,\ n=\dfrac{16}{1/4}=64$ ✓
• (C) Mean $20$, var $16$: $q=\tfrac{16}{20}=\tfrac45,\ p=\tfrac15,\ n=\dfrac{20}{1/5}=100$ ✓ …

Step 1 — Set up the binomial model

Each bulb is defective with probability $p=0.2$ independently; $n=5$ trials.

Step 2 — Apply the binomial formula for $X=1$

Step 3 — Compute

$(0.8)^4=0.4096,\qquad P=5\times0.2\times0.4096=1\times0.4096=0.4096.$ …

Step 1 — Identify the model

Each of 5 True-False questions is an independent trial with $p=\tfrac12$ of a correct guess, so $X=$ number correct is Binomial.

Step 2 — Substitute $n=5,\ r=3,\ p=\tfrac12$

Step 3 — Evaluate

$P(X=3)=\dfrac{10}{32}=0.3125$ …

Step 1 — The mode formula for a binomial

Step 2 — Substitute n = 7, p = 1/3

Since $2.67$ is not an integer, take its integer part: mode $=2$.

Why the other options are wrong: 3 would be $\lceil 8/3\rceil$; 7/3 and 8/3 are the mean $np=7/3$ and the value $(n+1)p=8/3$ respectively — neither is the mode, which must be a whole-number value the variable can actually take. …

We are drawing two cards with replacement from a standard 52‑card pack. Because we replace the card after the first draw, the two draws are independent and the probability of drawing an ace stays constant at $\frac{4}{52} = \frac{1}{13}$ each time.

The random variable $X$ counts the number of aces drawn in these two independent trials. This is exactly a binomial setting:

• Number of trials $n = 2$
• Probability of success (ace) on each trial $p = \frac{1}{13}$
• The trials are independent (thanks to replacement).

For any binomial distribution, the mean (expected value) is given by the simple formula

That’s the key concept: the mean of a binomial is just the number of trials times the success probability. No need to build the full probability distribution.

1. Identify $n$ and $p$:

   $n = 2$ (two draws), $p = \frac{1}{13}$.

2. Apply the formula:

3. Check the options: (A) $2$ — that would be the mean if every draw were an ace, impossible. (B) $\frac{2}{13}$ — matches our calculation. (C) $1$ — would be the mean if $p = \frac12$, not the case. (D) $\frac{1}{13}$ — that’s the mean for a single draw, not two. …

We are told that $X$ is a binomial variate, so $X \sim \text{Bin}(n, p)$.

For a binomial distribution:

• Mean = $np$
• Variance = $np(1-p)$

Given:

1. Find $p$ using the ratio of variance to mean

So $1-p = \frac{3}{5}$, hence

2. Find $n$ using the mean

So $X \sim \text{Bin}(8, \frac{2}{5})$.

3. Express $P(X > 1)$ in terms of $p$ and $n$

For binomial:

Here $p = \frac{2}{5}$, $1-p = \frac{3}{5}$, $n=8$.

So:

4. Combine the subtracted terms Factor $\left(\frac{3}{5}\right)^7$: $$$$

Since $A$ and $B$ are independent, $P(A\cap B)=P(A)\,P(B)=\dfrac25\cdot\dfrac13=\dfrac{2}{15}$, and each complement/conditional simplifies via independence.

List-I A) $P(\overline{A}\cup B)$

List-I B) $P(A\mid B)$ (independence $\Rightarrow$ conditioning has no effect)

List-I C) $P(A\cup B)$

List-I D) $P(\overline{B}\mid A)$

$P(\overline B\mid A)=P(\overline B)=1-\dfrac13=\dfrac23\quad(\text{I}).$ …

The problem gives us a person who gets success $\alpha$ times out of $\beta$ trials. That means the empirical probability of success in a single trial is $p = \frac{\alpha}{\beta}$. Consequently, the probability of failure in a single trial is $q = 1 - p = 1 - \frac{\alpha}{\beta} = \frac{\beta - \alpha}{\beta}$.

Now the experiment consists of $n$ independent trials. We want the probability that the person fails at least $(n-1)$ times. "At least $(n-1)$ failures" in $n$ trials means either:

• all $n$ trials are failures (0 successes), or
• exactly $n-1$ trials are failures (1 success).

These are the only two cases that satisfy the condition. So we need $P(\text{0 successes}) + P(\text{1 success})$.

1. Probability of 0 successes (all failures) Each trial fails with probability $q$, and all $n$ are independent:

2. Probability of exactly 1 success Choose which one of the $n$ trials is the success: $\binom{n}{1} = n$ ways. That one trial succeeds with probability $p$, and the other $n-1$ fail with probability $q^{n-1}$:

3. Add the two probabilities

Factor out the common term $\left(\frac{\beta - \alpha}{\beta}\right)^{n-1}$:

Combine the terms inside the bracket over the common denominator $\beta$:

$P = \left(\frac{\beta - \alpha}{\beta}\right)^{n-1} \cdot \frac{(\beta - \alpha) + n\alpha}{\beta}$ …

In problems involving a fixed number of independent trials, where each trial has only two possible outcomes (success or failure), we use the binomial probability distribution. This distribution helps us calculate the probability of getting a specific number of successes in a given number of trials.

The core idea here is to:

1. Determine the individual probabilities of success ($p$) and failure ($q$) for a single trial. The problem gives us a relationship between these two.
2. Identify the number of trials ($n$) and the specific event whose probability we need to find.
3. Apply the binomial probability formula for each component of the event and sum them up if the event comprises multiple possibilities.

Let's break down the solution step-by-step.

1. Determine the probabilities of success and failure for a single trial.

   Let $p$ be the probability of getting a success in a trial, and $q$ be the probability of getting a failure in a trial.

   We are given that the probability of getting a success is five times that of a failure:

   $p = 5q$

   We also know that for any trial, the sum of the probabilities of all possible outcomes must be 1. Since there are only two outcomes (success or failure):

   $p + q = 1$

   Now, substitute $p = 5q$ into the second equation:

   $5q + q = 1$

   $6q = 1$

   $q = \frac{1}{6}$

   Now find $p$:

   $p = 5q = 5 \times \frac{1}{6} = \frac{5}{6}$

   So, the probability of success is $\frac{5}{6}$ and the probability of failure is $\frac{1}{6}$.

2. Identify the parameters for the binomial distribution.

   We have $n=5$ trails.

   The probability of success in a single trial is $p = \frac{5}{6}$.

   The probability of failure in a single trial is $q = \frac{1}{6}$.

   The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability formula:

   $P(X=k) = \binom{n}{k} p^k q^{n-k}$

   where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

3. Define the event "at most one success".

   "At most one success" means the number of successes, $X$, can be either 0 or 1.

   So, we need to calculate $P(X \le 1) = P(X=0) + P(X=1)$. …

The question is about the variance of the number of successes when a die is rolled 5 times, where "success" means getting an odd number. This is a classic binomial distribution problem. The key is to recognize that each roll is independent and has the same probability of success — the die has 6 faces, 3 of which are odd (1, 3, 5), so the probability of success on a single trial is $p = \frac{3}{6} = \frac{1}{2}$.

For a binomial distribution with $n$ trials and success probability $p$, the variance is given by $np(1-p)$. This formula comes from the fact that each trial contributes $p(1-p)$ to the variance, and trials are independent, so variances add. No need to derive from scratch — just apply it.

Let’s walk through it step by step.

1. Identify the distribution. Rolling a die 5 times, counting successes (odd numbers), gives a binomial distribution with parameters $n = 5$ and $p = \frac{1}{2}$. Each trial has exactly two outcomes: success (odd) or failure (even), with constant probability.

2. Recall the variance formula for a binomial distribution. For a binomial random variable $X \sim \text{Binomial}(n, p)$, the variance is $\text{Var}(X) = np(1-p)$. This is a standard result — it comes from the fact that $X$ is the sum of $n$ independent Bernoulli variables, each with variance $p(1-p)$. …