Question 1035228
A binomial experiment has 5 trials in which p = 0.7. What is the probability of getting at least 2 successes?
my equation I came up with is p(0)+p(1)=5c0(.4)^0(.6)^5-0+5c1(.4^1(.6)^5-0
<pre>I don't know where you got .4 from. The probability of SUCCESS (p) is .7, not .4.
You also had the right idea, it appears, in determining the probability of getting @ MOST 1, which is: 
{{{" "[5]C[0] (.7)^0 (.3)^(5 - 0) + " "[5]C[1] (.7)^1 (.3)^(5 - 1)}}}, but then to get @ least 2, which is the SUM of P(2) through P(5), 
cumulative, you'd need to subtract this probability from 1.
We then get: {{{1 - (" "[5]C[0] (.7)^0 (.3)^(5 - 0) + " "[5]C[1] (.7)^1 (.3)^(5 - 1))}}} = 1 - .0308 = {{{highlight_green(.9692)}}}

If you're familiar with MS Excel, you could use 1 - BINOMDIST to get this probability
If you're familiar with the TI-83/84 calculator, you could use: BINOMCDF to get this probability