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A.$\dfrac{1}{6}$

B.$\dfrac{1}{4}$

C.$\dfrac{1}{8}$

D.$\dfrac{1}{3}$

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Hint:It is given in question that, $4P(X=4)=P(X=2)$ and $n=6$ . Apply the formula $P(X=x)={}^{n}{{C}_{x}}{{p}^{n}}{{q}^{1-n}}$ and solve it, you will get the answer.

Complete step-by-step Solution:

In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of $n$ independent experiments, each asking a yes–no question, and each with its own boolean valued outcome: success/yes/true/one or failure/no/false/zero (with probability q = 1 − p). A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., $n=1$ , the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely it is that a proposition is true. Probability is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the two outcomes ("heads" and "tails") are both equally probable; the probability of "heads" equals the probability of "tails"; and since no other outcomes are possible, the probability of either "heads" or "tails" is 1/2 (which could also be written as 0.5 or 50%).

The literal meaning of Independent Events is the events which occur freely of each other. The events are independent of each other. In other words, the occurrence of one event does not affect the occurrence of the other. The probability of occurring of the two events are independent of each other.

An event A is said to be independent of another event B if the probability of occurrence of one of them is not affected by the occurrence of the other.

Suppose if we draw two cards from a pack of cards one after the other. The results of the two draws are independent if the cards are drawn with replacement i.e., the first card is put back into the pack before the second draw. If the cards are not replaced then the events of drawing the cards are not independent.

It is given in question that, $4P(X=4)=P(X=2)$ and $n=6$ .

$4P(X=4)=P(X=2)$

$\begin{align}

& 4{}^{6}{{C}_{4}}{{p}^{4}}{{q}^{2}}={}^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}} \\

& 4.\dfrac{6\times 5\times 4\times 3}{4\times 3\times 2\times 1}{{p}^{4}}{{q}^{2}}=\dfrac{6\times 5}{2\times 1}{{p}^{2}}{{q}^{4}} \\

& 4.15{{p}^{4}}{{q}^{2}}=15.{{p}^{2}}{{q}^{4}} \\

& 4{{p}^{2}}={{q}^{2}} \\

\end{align}$

We know, $p=1-q$,

$\begin{align}

& 4{{p}^{2}}={{(1-p)}^{2}} \\

& 4{{p}^{2}}=1+{{p}^{2}}-2p \\

& 3{{p}^{2}}+2p-1=0 \\

\end{align}$

Now solving the quadratic equation we get,

\[\begin{align}

& 3{{p}^{2}}+2p-1=0 \\

& 3{{p}^{2}}+3p-p-1=0 \\

& 3p(p+1)-1(p+1)=0 \\

& (3p-1)(p+1)=0 \\

\end{align}\]

\[(3p-1)(p+1)=0\]

We get,

\[p=\dfrac{1}{3},-1\]

The negative value cannot be taken because probability can’t be negative .

\[p=\dfrac{1}{3}\]

We get the value of \[p=\dfrac{1}{3}\] .

The correct answer is option (D).

Note: Read the question carefully. Your concept regarding the probability should be clear. Also, do not confuse yourself. Take utmost care that no terms are missing. Avoid silly mistakes. While simplifying do not jumble yourself.

Complete step-by-step Solution:

In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of $n$ independent experiments, each asking a yes–no question, and each with its own boolean valued outcome: success/yes/true/one or failure/no/false/zero (with probability q = 1 − p). A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., $n=1$ , the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely it is that a proposition is true. Probability is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the two outcomes ("heads" and "tails") are both equally probable; the probability of "heads" equals the probability of "tails"; and since no other outcomes are possible, the probability of either "heads" or "tails" is 1/2 (which could also be written as 0.5 or 50%).

The literal meaning of Independent Events is the events which occur freely of each other. The events are independent of each other. In other words, the occurrence of one event does not affect the occurrence of the other. The probability of occurring of the two events are independent of each other.

An event A is said to be independent of another event B if the probability of occurrence of one of them is not affected by the occurrence of the other.

Suppose if we draw two cards from a pack of cards one after the other. The results of the two draws are independent if the cards are drawn with replacement i.e., the first card is put back into the pack before the second draw. If the cards are not replaced then the events of drawing the cards are not independent.

It is given in question that, $4P(X=4)=P(X=2)$ and $n=6$ .

$4P(X=4)=P(X=2)$

$\begin{align}

& 4{}^{6}{{C}_{4}}{{p}^{4}}{{q}^{2}}={}^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}} \\

& 4.\dfrac{6\times 5\times 4\times 3}{4\times 3\times 2\times 1}{{p}^{4}}{{q}^{2}}=\dfrac{6\times 5}{2\times 1}{{p}^{2}}{{q}^{4}} \\

& 4.15{{p}^{4}}{{q}^{2}}=15.{{p}^{2}}{{q}^{4}} \\

& 4{{p}^{2}}={{q}^{2}} \\

\end{align}$

We know, $p=1-q$,

$\begin{align}

& 4{{p}^{2}}={{(1-p)}^{2}} \\

& 4{{p}^{2}}=1+{{p}^{2}}-2p \\

& 3{{p}^{2}}+2p-1=0 \\

\end{align}$

Now solving the quadratic equation we get,

\[\begin{align}

& 3{{p}^{2}}+2p-1=0 \\

& 3{{p}^{2}}+3p-p-1=0 \\

& 3p(p+1)-1(p+1)=0 \\

& (3p-1)(p+1)=0 \\

\end{align}\]

\[(3p-1)(p+1)=0\]

We get,

\[p=\dfrac{1}{3},-1\]

The negative value cannot be taken because probability can’t be negative .

\[p=\dfrac{1}{3}\]

We get the value of \[p=\dfrac{1}{3}\] .

The correct answer is option (D).

Note: Read the question carefully. Your concept regarding the probability should be clear. Also, do not confuse yourself. Take utmost care that no terms are missing. Avoid silly mistakes. While simplifying do not jumble yourself.

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