Given that, a pair of dice is thrown 4 times.

And a doublet is considered as a success.

Let p be the probability of getting a doublet in a throw of a pair of dice.

Since, there are 36 possible outcomes in total. {(1, 1), (1, 2), (1, 3), …, (1, 6), ..., (2, 6), …, (6 ,6)}

And the 6 possible doublets in 36 outcomes. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

So,

And also, p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing a number of doublets (successes) out of 4 throws.

So, Binomial distribution of getting r successes out of 4 throws is given by

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 4.

Now, substituting values of n, p and q in the formula P (X = r). We get

…(i)

We need to find the probability distribution of the number of successes.

The probability of 0 success in 4 throws is given by,

Probability = P (X = 0)

Put r = 0 in (i),

The probability of 1 success in 4 throws is given by,

Probability = P (X = 1)

Put r = 1 in (i),

The probability of 2 successes in 4 throws is given by,

Probability = P (X = 2)

Put r = 2 in (i),

The probability of 3 successes in 4 throws is given by,

Probability = P (X = 3)

Put r = 3 in (i),

The probability of 4 successes in 4 throws is given by,

Probability = P (X = 4)

Put r = 4 in (i),

Thus, the probability distribution is