Bernoulli Trial Help!

Question

Assume that $n = 9$, and $p = \frac{4}{5}$ .

Find the probability of at least 3 successes and at least 2 failures.

What I have so far:

$c(9,3)\cdot (\frac{4}{5})^3\cdot (\frac{1}{5})^6=.002753$

$c(9,4)\cdot (\frac{4}{5})^4 \cdot (\frac{1}{5})^5=.016515$

$c(9,5)\cdot (\frac{4}{5})^5\cdot (\frac{1}{5})^4=.06606$

This is about as far as I have been able to get. I get really thrown off at the at least 2 failures part. I think I need to find the chances of 2 failures and add it to the numbers above but am unsure where to put my numbers. Or I could be completely wrong with that assumption. Any thoughts? Thanks in advance!

Answer 1

Let X follow Binomial(n,p). Then your answer is P(3<= X <= 7) which is the summation of the terms: (nCx)(p^x)((1-p)^(n-x)) where x ranges from 3 to 7.

Answer 2

You have to calculate $p\{3\leq X\leq 7\}$ because to have at least $2$ failure is equivalent to have at most $7$ success.

Then, you have to calculate $$p\{3\leq X\leq 7\}=\sum_{k=3}^7\binom{9}{k}p^k(1-p)^{9-k}$$