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What is the probability distribution when a fair die is rolled once?

Mathematics

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Answer

P(X=x)=16 for x∈{1,2,3,4,5,6}P(X=x) = \frac{1}{6} \text{ for } x \in \{1, 2, 3, 4, 5, 6\}

Step-by-step solution

4 steps
  1. Identify the sample space.

    When a fair die is rolled once, the possible outcomes for the up-turned face are the integers from 1 to 6.
    The sample space S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}.
    The total number of possible outcomes is 6.

  2. Define the random variable.

    Let XX be the random variable representing the number on the up-turned face. The possible values for XX are x∈{1,2,3,4,5,6}x \in \{1, 2, 3, 4, 5, 6\}.

  3. Calculate the probability for each outcome.

    Since the die is fair, each outcome has an equal probability of occurring.
    The probability of any specific outcome xx is given by:

    P(X=x)=Number of favorable outcomesTotal number of outcomesP(X=x) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

    For each value of xx from 1 to 6, there is only one favorable outcome (e.g., rolling a 1, rolling a 2, etc.).
    Therefore, for each x∈{1,2,3,4,5,6}x \in \{1, 2, 3, 4, 5, 6\}:

    P(X=x)=16P(X=x) = \frac{1}{6}
  4. Present the probability distribution.

    The probability distribution can be shown in a table:

    xP(X=x)116216316416516616\begin{array}{|c|c|} \hline x & P(X=x) \\ \hline 1 & \frac{1}{6} \\ 2 & \frac{1}{6} \\ 3 & \frac{1}{6} \\ 4 & \frac{1}{6} \\ 5 & \frac{1}{6} \\ 6 & \frac{1}{6} \\ \hline \end{array}

    The sum of all probabilities is 16×6=1\frac{1}{6} \times 6 = 1, which confirms it is a valid probability distribution.

    The probability distribution for the up-turned faces is:

    P(X=x)=16 for x∈{1,2,3,4,5,6}P(X=x) = \frac{1}{6} \text{ for } x \in \{1, 2, 3, 4, 5, 6\}
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