1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – whuber Apr 15 '13 at 16:57
2
$\begingroup$

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, $$P(\text{no}~2~\text{shows}) = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \Rightarrow P(\text{at least one}~2) = 1-\frac{25}{36}=\frac{11}{36}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.