Even though this lesson is about probability, my question is about algebra. Would someone explain to me, or show me the steps, to getting 1=((1/6)+(5/6))^3, which appears to be in the form 1=(a+b)^3, by adding up the numbers above it? I keep doing the math wrong. Thanks.
View attachment 8988
This question comes from this video about 13:04 minutes in,
https://www.youtube.com/watch?v=zc9QKwbuUQk
You are rolling 1 fair die three times. What is the probability of rolling a 5 on a single roll? 1/6. So what is the probability of rolling anything other than 5? 1 - 1/6 = 5/6.
Probabilty of no 5's is
\(\displaystyle \left ( \dfrac{5}{6} \right )^3 = 1 * 1 * \left ( \dfrac{5}{6} \right ) = \dbinom{3}{0} * \left ( \dfrac{1}{6} \right )^0 * \left ( \dfrac{5}{6} \right )^3.\)
Buy that?
There are 3 ways to get ONE 5, on the first, second, or third roll. So probability of exactly one 5 is
\(\displaystyle 3 * \dfrac{1}{6} * \left ( \dfrac{5}{6} \right )^2 = \dbinom{3}{1} * \left ( \dfrac{1}{6} \right )^1 * \left ( \dfrac{5}{6} \right )^2.\)
Make sense?
There are 3 ways to get TWO 5's, first and second, first and third, or second and third. So probability of exactly two is
\(\displaystyle 3 * \left ( \dfrac{1}{6} \right )^2 * \dfrac{5}{6} = \dbinom{3}{2} * \left ( \dfrac{1}{6} \right )^2 * \left ( \dfrac{5}{6} \right )^3.\)
Still with me?
And what about THREE 5's?
\(\displaystyle \left ( \dfrac{1}{6} \right )^3 = \dbinom{3}{3} * \left ( \dfrac{1}{6} \right )^3 * \left ( \dfrac{5}{6} \right )^0.\)
Straight forward. Now add them up to get.
\(\displaystyle \displaystyle \sum_{j=0}^3 \dbinom{3}{j} \left ( \dfrac{1}{6} \right )^j * \left ( \dfrac{5}{6} \right)^{(3-j)} = \left ( \dfrac{1}{6} + \dfrac{5}{6} \right )^3 = 1^3 = 1.\)
