Question

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{4}{7}$

$\dfrac{5}{7}$

$\dfrac{3}{4}$

Answer: C

Concepts:complementary counting cube geometry combinations

Difficulty rating: 1860

Solution:

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the $6$ faces gives $\binom{4}{3} = 4$ triples, so $6 \cdot 4 = 24$ triples lie on a face out of $\binom{8}{3} = 56$ total.

The probability of hitting the interior is $1 - \dfrac{24}{56} = \dfrac{4}{7}.$

Thus, the correct answer is C.

2009 AMC 10A Problem 24

Answer: C

Difficulty rating: 1860

Solution:

Problem 24 in Other Years