find the probability distribution of the number of doublets in 4 throws of a pair of dice.

Here is the answer for your question 

Probability of getting doublet in single throw =6/36 = 1/6

  Probability of not getting doublet in single throw = 30/36 = 5/6

 Let X denote a random variable which is the number of doublets obtained  in four throws of a pair of dice.

X  has the binomial distribution with n = 4.

P(X = x)  = ⁿC_x p^x q^{n - x} 

P(X = 0)  = ⁴C₀ p⁰ q⁴   = (5/6)⁴ = 625/1296

P(X = 1)  = 4 (1/6)¹ (5/6)³ = 500/1296

P(X = 2)  = 6 (1/6)² (5/6)²  = 150/1296

P(X = 3)  = (1/6)⁴ = 1/1296