Question

multiple choice 2 points
the four faces of a fair tetrahedron die are numbered 1, 2, 3, and 4. in two tosses of
the die, the probability of obtaining exactly two 4s is
$\frac{1}{16}$
$\frac{1}{2}$
$\frac{1}{8}$
$\frac{1}{4}$

Explanation:

Step1: Calculate total outcomes

When a tetrahedron die is tossed twice, the total number of outcomes is \(n(S)=4\times4 = 16\) (by the fundamental counting principle).

Step2: Calculate favorable outcomes

The favorable outcomes (exactly two 4's) is \(n(E)=1\) (i.e., \((4,4)\)).

Step3: Calculate probability

The probability formula is \(P(E)=\frac{n(E)}{n(S)}\). Substituting the values, we get \(P(E)=\frac{1}{16}\).

Answer:

\(\frac{1}{16}\)