How to calculate the probility of 2 independent events of having same value? We are learning to calculate the probability of sums and difference of random numbers.
Here is the problem:
One athlete knows from past experience that the distances of his javelin throws follow a normal distribution with mean 60 (units) and standard deviation 6 (units).  The distances of his hammer throws follow a normal distribution with mean 54 (units) and standard deviation 4 (units).  The athlete's performances in the two events are independent.
I know how to calculate probability if his next javelin throw is smaller/greater than hammer throw.
I can calculate probability of both javelin/hammer throw is greater or smaller than some distance.
But how to calculate the probability that events (javelin throw and hammer throw) can have same value?
to solve the we are using z test with the statistics table for z values.
*I also changed values if you want you can change again.
 A: The probability that two independent continuous random variables
are exactly equal is 0. And, of course, normal random variables are continuous. Continuous random variables put 0 probability
on any one point.
A more interesting problem might be the probability that the two
variables are within one unit of each other.
For discrete random variables the answer need not be 0.
For example, If I roll two fair dice, the probability they
will show equal numbers is 1/6. 
Addendum (A day later): Here is a brief simulation in R. If
you are sufficiently interested in the probability that the two
random variables match within a unit or half a unit, you can
do the formal analysis and check results with the output of
the simulation. (With a million iterations, typically accurate to about 2 significant digits.) What is the distribution of $X - Y?$
The last result is an exact one.
 m = 10^6;  x = rnorm(m, 60, 6);  y = rnorm(m, 54, 4)
 mean(x == y)
 ## 0  # simulation actually discrete, but to enough places to avoid match 
 mean(abs(x - y) < 1)
 ## 0.078427  # approx. P{|X-Y| < 1}
 mean(abs(x - y) < .5)
 ## 0.039537

 # An EXACT computation:
 mu.dif = 60-54;  sd.dif = sqrt(6^2 + 4^2)
 diff(pnorm(c(-1,1), mu.dif, sd.dif)) # 'pnorm' is CDF
 ## 0.07819451

