Example (TI83): First, we will find 5 factorial. 

To find 5! using
the 

We get the answer 120. 

Example (TI83): Find the probability that 2 successes will occur
out of 5 trials if the probability of success is 0.3. That is, find P(x = 2). 

Press 2^{nd} VARS [DISTR]. Scroll down to A:binompdf( Press ENTER. 

Enter 5,.3,2) and press ENTER to get the answer .3087. The syntax is binompdf(n, p, xvalue). 

Example (TI83): Find the probability that 2 or less successes will
occur out of 5 trials if the probability of success is 0.3. That is, find P(x
≤ 2). 

Press 2^{nd} VARS [DISTR]. Scroll down to B:binomcdf( Press ENTER. 

Enter 5,.3,2) and press ENTER to get the answer .83692. The syntax is binomcdf(n, p, largest xvalue). 

Example (TI83): Find the probability that at least 2 successes
will occur out of 5 trials if the probability of success is 0.3. That is,
find P(x ≥ 2). Use the complement rule to find P(x ≥ 2) = 1 – P(x ≤ 1). 

Press 1 – and 2^{nd} VARS [DISTR]. Scroll down to B:binomcdf( Press ENTER. 

Enter 5,.3,1) and press ENTER to get the answer .47178. The syntax is 1 – binomcdf(n, p, largest xvalue). 