Question

with a fair die, the probability of rolling any number 1 through 6 is the same. it $p(1)$ represents the probability of rolling a 1, $p(2)$ the probability of rolling a 2, and so forth, what is the value of: $p(1)+p(2)+p(3)+p(4)+p(5)+p(6)$?

Explanation:

Step1: Determine probability of single event

A fair die has 6 possible outcomes (1 - 6), each with equal probability. So \( P(1)=P(2)=\dots = P(6)=\frac{1}{6} \).

Step2: Sum the probabilities

We need to calculate \( P(1)+P(2)+P(3)+P(4)+P(5)+P(6) \). Substituting the value of each probability: \( \frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \). There are 6 terms of \( \frac{1}{6} \), so the sum is \( 6\times\frac{1}{6}=1 \). Also, by the definition of probability, the sum of probabilities of all possible outcomes in a sample space is 1.

Answer:

1