# Finding the bionomial probability?

Ok so i have been taught this formula regarding binomial probability

Repeat an even n times

  x = # of successes
p = probability of successes
q = probability of failure


p(x = a) = nCa * p^a * q^(n-a)

Ok wow seems good. So i try to solve this question using this formula.

A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one 2.

What is the probability of seeing at least one 2 in any one roll of the pair of dice?

The probability of seeing at least one 2 = 1 - probability of seeing no 2s at all

So i calculate the probability of seeing no 2s at all.

x = 0 p = 1/6 q = 5/6 n = 10

Using the formula, p(x=0) = 10C0 * (1/6)^0 * (5/6)^10 Which gives me, p(x=0) = 0.1615

Probability of seeing at least one 2 = 1 - 0.1615 = 0.83

And i checked the ans, it stated the ans was 0.306.

So where did i go wrong? If someone can help me please, i would be much grateful

• The two lines leading up to the question ("A pair..." and "Let X...") seem not to have any relevance to the question itself. Is something perhaps missing in its statement? – whuber Apr 15 '13 at 16:57

The probability of seeing at least one $2$ when you roll a pair of fair dice is \begin{align}P(\text{at least one}~2)&=P(\text{first shows}~2)+P(\text{second shows}~2)-P(\text{both show}~2)\\&=\frac{1}{6}+\frac{1}{6}-\left(\frac{1}{6}\right)^2\\&=\frac{11}{36}.\end{align} Alternatively, $$P(\text{no}~2~\text{shows}) = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \Rightarrow P(\text{at least one}~2) = 1-\frac{25}{36}=\frac{11}{36}.$$