Easier way for probability problem? 
Question: Bob takes a math test with $10$ questions. He has a $70\%$ chance of getting each question right. What is the probability that they get exactly $60\%$ on the test? Express your answer as a common fraction.

My bashy way: We know that the successful outcomes (numerator of probability) is $\binom{10}{6} \left(\frac{7}{10}\right)^6 \left(\frac{3}{10}\right)^4$ since we choose $6$ to be correct and the rest $4$ to be wrong. Following in vein, the denominator of our probability is just
$$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}.$$
Is there a slicker way to do this probability problem?
 A: We know that the successful outcomes (numerator of probability)
look here: Binomial distribution.
The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function: $$\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}$$
Therefore your score is the exact probability of 6 successes in 10 independent trials. Following:
$$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=\Pr(X=1)+\Pr(X=2)+...+\Pr(X=3)=\Pr(X\ge1)$$
the result of which the probability that Bob has a result greater than or equal to 10%. So that Bob will answer at least one question correctly. Look: 0.999994 which is the opposite of the fact that Bob will answer each question wrong: $$\Pr(\text{all wrong})=\Pr(X=0)=\left(\frac{3}{10}\right)^{10}$$ $$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1-\Pr(\text{all wrong})$$ $$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1-\left(\frac{3}{10}\right)^{10}$$ $$\left(\frac{3}{10}\right)^{10}+\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1$$ $$\sum_{n=0}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1$$
