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Eugene has three six‐sided dice that he rolls simultaneously. The faces of the dice
contain these letters:
Die #1: A B C D E F
Die #2: G H I J K L
Die #3: M N O P Q R
a. What is the probability that he will roll three vowels?



Answer :

Answer:

The probability that he will roll three vowels = [tex]\displaystyle\bf\frac{1}{108}[/tex]

Step-by-step explanation:

We can find the probability that he will roll three vowels by using the probability formula:

[tex]\boxed{P(A)=\frac{n(A)}{n(S)} }[/tex]

where:

  • [tex]P(A)[/tex] = probability of event A
  • [tex]n(A)[/tex] = number of outcomes of event A
  • [tex]n(S)[/tex] = number of total outcomes

Let:

A = rolling a vowel on Die #1

  = {A, E}

n(A) = 2

n(S) = 6

[tex]\displaystyle P(A)=\frac{n(A)}{n(S)}[/tex]

[tex]\displaystyle P(A)=\frac{2}{6}[/tex]

[tex]\displaystyle P(A)=\frac{1}{3}[/tex]

Let:

B = rolling a vowel on Die #2

  = {I}

n(B) = 1

n(S) = 6

[tex]\displaystyle P(B)=\frac{n(B)}{n(S)}[/tex]

[tex]\displaystyle P(B)=\frac{1}{6}[/tex]

Let:

C = rolling a vowel on Die #3

  = {O}

n(C) = 1

n(S) = 6

[tex]\displaystyle P(C)=\frac{n(C)}{n(S)}[/tex]

[tex]\displaystyle P(C)=\frac{1}{6}[/tex]

Since event A, B, and C are independent events. Then the probability of rolling three vowel:

[tex]P=P(A)\times P(B)\times P(C)[/tex]

   [tex]\displaystyle=\frac{1}{3} \times\frac{1}{6} \times\frac{1}{6}[/tex]

   [tex]\displaystyle=\bf\frac{1}{108}[/tex]

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