The corners of regular tetrahedrons are numbe | vedclass

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(c) Required combinations are $(2, 2, 1), (1, 2, 2), (2, 1, 2), (1, 3, 1,), (3, 1, 1)$ and $(1, 1, 3)$

$	herefore$ Required probability $ = \

Vedclass · Accepted Answer

$\frac{3}{{32}}$

Vedclass · Answer

(c) Required combinations are $(2, 2, 1), (1, 2, 2), (2, 1, 2), (1, 3, 1,), (3, 1, 1)$ and $(1, 1, 3)$

$	herefore$ Required probability $ = \frac{6}{{{4^3}}} = \frac{6}{{64}} = \frac{3}{{32}}$.

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

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