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14.Probability
medium
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
$\frac{3}{{16}}$
Solution
(c) Required combinations are $(2, 2, 1), (1, 2, 2), (2, 1, 2), (1, 3, 1,), (3, 1, 1)$ and $(1, 1, 3)$
$\therefore$ Required probability $ = \frac{6}{{{4^3}}} = \frac{6}{{64}} = \frac{3}{{32}}$.
Standard 11
Mathematics
