The corners of regular tetrahedrons are numbe | vedclass

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(C) The total number of outcomes when tossing three tetrahedrons is $4^3 = 64$.We need to find the number of outcomes where the sum of the upward corn

Vedclass · Accepted Answer

$\frac{3}{32}$

Vedclass · Answer

(C) The total number of outcomes when tossing three tetrahedrons is $4^3 = 64$.We need to find the number of outcomes where the sum of the upward corners is $5$.Let the outcomes be $(x, y, z)$ where $x, y, z \in \{1, 2, 3, 4\}$.The possible combinations that sum to $5$ are:$(1, 1, 3)$ (can occur in $3$ permutations: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$)$(1, 2, 2)$ (can occur in $3$ permutations: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$)Total favorable outcomes = $3 + 3 = 6$.Therefore,the required probability = $\frac{6}{64} = \frac{3}{32}$.

The corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is

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