Compute the probability PX 2 if the random variable X has a

Compute the probability P(X 2) if the random variable X has a Poisson distribution satisfying P(X = 4) = P(X = 5).

Possion Distribution
PMF of P.D is = f ( k ) = e- ^x / x!
Where
= parameter of the distribution.
x = is the number of independent trials

For mean =

P( X = 5 ) = e ^- * ^5 / 5!
P( X = 4 ) = e ^- * ^4 / 4!

It is given that P( X = 4 ) = P( X = 5 )
e ^- * ^5 / 5! = e ^- * ^4 / 4!
=> ^5 / 5! = ^4 / 4!
=> ^5 / ^4 = 5!/ 4!
=> = 5 * 4!/ 4!
=> = 5

mean = = 5

P( X > = 2 ) = 1 - P (X < 2)

P( X < 2) = P(X=1) + P(X=0)
= e^-5 * 4 ^ 1 / 1! + e^-5 * ^ 0 / 0!
= 0.0404
P( X > = 2 ) = 1 - P (X < 2) = 0.9596