Let x be a random variable with the following probability distribution:

X	0	1	2	3
P (x)	0.4	0.3	0.2	0.1

Does x have a binomial distribution? Explain your answer. Please show solution.

The answer is not

n = 3 , np = 1 => p = 1/3 and q = 2/3

if x = 0 => p [x = 0} = 0.4

if x = 1 => p [ x = 1 ] = 0.3

if x = 2 => p [x = 2 ] = 0.2

if x = 3 => p [ x = 3 ] = 0.1 BECAUSE...

Note that the distribution for n = 3, p = 1/3 looks like. P(x = 0) = C(3,0) (1/3) ^ 0 * (2/3) ^ 3 =8/27 P(x = 1) = C(3,1) * (1/3) ^ 1 * (2/3) ^ 2 = 12/27 P(x = 2) = C(3,2) * (1/3) ^ 2 * (2/3) ^ 1 = 6/27 P(x = 3) = C(3,3) * (1/3) ^ 3 * (2/3) ^ 0 = 1/27

since this different from the given distribution, the given distribution is therefore not the binomial distribution with n = 3 and p = 1/3.

I believe it is not fit for a binomial distribution but HOW? Please show solution and explain answer.