# Binomial Distribution Question - probability summations

Pi Dumbo is not very bright, but he makes up for it by playing video games all day. He sits for a 20-question multiple choice examination, each question having five possible answers. He fills in the answers on the ubble sheet randomly without bothering to read the questions. Let Y denote the number of questions that Pi Dumbo answers correctly. Find $\Pr[Y>2]$.

I calculate the probability distribution: $\Pr[Y=y]=\binom{20}{y} \cdot 0.2^y \cdot 0.8^{20-y}$

To solve the question, I thought I would do the following:

$\Pr[Y>2] = 1 - \Pr[Y \le 2] = 1- \sum\limits_{i=0}^{2}\Pr[y_i]\tag{1}$.

I get the wrong answer doing this. Apparently, the upper limit on the summation for equation (1) should be 1. But this doesn't make sense to me according to the 3 laws of probability. Is there a typo in my book?

I did another question a similar way and got the right answer:

Kobe Bryant is an eighty percent free-thrower shooter. In a game against the Portland Trailbrazers, Kobe went to the foul line twelve times. Let $M$ denote the number of free-throws that Kobe makes. Assume his shots are independent. Calculate:

(a) the probability distribution,

(b) $E[M]$

(c) $Var[M]$

(d) What is the probability that Kobe misses more than two shots?

$\Pr[M=m]=\binom{12}{m} \cdot 0.8^m \cdot 0.2^{12-m}$

$E[M] = 12*0.8=9.6$

$Var[M]=12*0.8*0.2=1.92$

(d) $\Pr[12-M>2]=\Pr[M<10]=1-\Pr[M \ge 10]=0.4417$

If it says $\gt 2$ then your procedure is correct. Note that we do have to take $0$ into account, people often forget it. – André Nicolas Jul 17 '13 at 1:35