Finding the bionomial probability? - Cross Validated most recent 30 from stats.stackexchange.com 2019-07-16T06:52:48Z https://stats.stackexchange.com/feeds/question/56169 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/56169 1 Finding the bionomial probability? kype https://stats.stackexchange.com/users/24311 2013-04-15T16:43:23Z 2013-04-15T17:40:16Z <p>Ok so i have been taught this formula regarding binomial probability</p> <blockquote> <p>Repeat an even n times </p> <pre><code> x = # of successes p = probability of successes q = probability of failure </code></pre> <p>p(x = a) = nCa * p^a * q^(n-a)</p> </blockquote> <p>Ok wow seems good. So i try to solve this question using this formula.</p> <blockquote> <p>A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one 2.</p> <p>What is the probability of seeing at least one 2 in any one roll of the pair of dice?</p> </blockquote> <p>The probability of seeing at least one 2 = 1 - probability of seeing no 2s at all</p> <p>So i calculate the probability of seeing no 2s at all.</p> <p>x = 0 p = 1/6 q = 5/6 n = 10</p> <p>Using the formula, p(x=0) = 10C0 * (1/6)^0 * (5/6)^10 Which gives me, p(x=0) = 0.1615</p> <p>Probability of seeing at least one 2 = 1 - 0.1615 = 0.83</p> <p>And i checked the ans, it stated the ans was 0.306.</p> <p>So where did i go wrong? If someone can help me please, i would be much grateful</p> https://stats.stackexchange.com/questions/56169/-/56173#56173 2 Answer by Dilip Sarwate for Finding the bionomial probability? Dilip Sarwate https://stats.stackexchange.com/users/6633 2013-04-15T16:58:39Z 2013-04-15T17:40:16Z <p>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)&amp;=P(\text{first shows}~2)+P(\text{second shows}~2)-P(\text{both show}~2)\\&amp;=\frac{1}{6}+\frac{1}{6}-\left(\frac{1}{6}\right)^2\\&amp;=\frac{11}{36}.\end{align} Alternatively, $$P(\text{no}~2~\text{shows}) = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \Rightarrow P(\text{at least one}~2) = 1-\frac{25}{36}=\frac{11}{36}.$$</p>