$\begin{array}{1 1} \large \frac{25}{216} \\\large \frac{5}{216} \\ \large \frac{25}{36} \\ \large \frac{5}{36}\end{array} $

- For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n - x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 - p)$
- Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.

The experiment is to toss a pair of dice (n = 4) times. Getting a doublet is considered a success. It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.

When a pair of dice is thrown once, the number of possible outcomes = 6 $\times$ 6 = 36.

Doublets are the same number appearing on both die.

The number of possible doublets in this case is : {(1,1), (2,2), (3,3), (4,4), (5,5), (6.6)}

P (getting a doublet on a roll of two dice) = $p \;=\;\large\frac{6}{36} = \frac{1}{6}$

Therefore P (not getting a doublet) = $q\;1-p=1-\large\frac{1}{6} =\frac{5}{6}$

X is a bionomial distribution. Here $n=4, p = \large\frac{1}{6}, $ $q =\large \frac{5}{6}$.

Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n - x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 - p)$

$\Rightarrow P (X=2 success) = ^{4}C_2 \times \large\frac{1}{6}^2 $$\times \large\frac{5}{6}^{4-2}$

$\Rightarrow P (X=2) = \large\frac{4 \times 3}{2}$$ \times \large \frac{1}{36} $$\times\large \frac{25}{36} = \frac{25}{216}$

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