# What is the idea of ‘probability p’?

Hello mathstackexchange

I’m working my way through a book on mathematics and have come to a section on probability. I’ve not done probability before and whilst the resource has been excellent, in the word problems section of the text, it is using some terminology I can’t translate into anything I know.

The expression in context that’s confusing me issues is “This die has a probability of p of obtaining a six. Answer parts i-iii in terms of p.

What is this “probability p”?

(The question refers to an earlier situation I have analyzed and answered questions about by constructing a tree diagram. The situation is a dice game where three six sided dice are rolled simultaneously and points are awarded based on the number of 6s that show up with a points bonus if three 6s are rolled. I’ve included a picture of the tree).

Posting the questions relating to the word-problem with the unknown terminology here for further context, not seeking answers or strategies on them: i) Determine probability of rolling no sixes. ii) Determine therefore the probability of scoring points in a round iii) If the probability of scoring max points in a round is $$1/27$$, what is the value of $$p$$? iv) If the probability of obtaining exactly $$16$$ points in a round is $$0.032$$ while the probability of obtaining exactly $$10$$ points is $$0.128$$, determine the value of $$p$$.

• $p$ here stands for a real number in interval $[0,1]$. Feb 7 at 22:22

Probability $$p$$ essentially refers to the idea of some event happening with chance $$p$$, where $$p \in [0,1]$$. If $$p=1$$ then it is certain to happen, if $$p=0$$ then it is impossible to happen. The dice example relies on all outcomes being equally likely, you'll cover this in detail as you progress through the book. But roughly, probability measures the likelihood of something happening.
• Looking at (iii) and (iv) I think in this problem the die may be loaded so you do not have to have $p=\frac16$ Feb 7 at 22:30
• Your claims about the cases $p=0,\,p=1$ aren't quite right in general. However, they'll suffice for the discrete probability distribution of a dice roll.
In your tree, you use $$\frac{1}{6}$$ for the probability of rolling a $$6$$, which is valid for a standard fair die. But what if the die is non-standard (perhaps with two sixes instead of one) or unfair (weighted so that the six come up more often)? We introduce the variable $$p$$ to represent the (possibly unknown) probability of rolling a six. For a standard fair die $$p=\frac{1}{6}$$, for a die with two sixes $$p=\frac{1}{3}$$ and for an unfair die we might have $$p=\frac{1}{5}$$ or any other value $$0\le p\le1$$.