The corners of regular tetrahedrons are numbered $1, 2, 3, 4.$ Three tetrahedrons are tossed. The probability that the sum of upward corners will be $5$ is
The corners of regular tetrahedrons are numbered $1, 2, 3, 4.$ Three tetrahedrons are tossed. The probability that the sum of upward corners will be $5$ is
- A
$\frac{5}{{24}}$
- B
$\frac{5}{{64}}$
- C
$\frac{3}{{32}}$
- D
$\frac{3}{{16}}$
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