Question

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Answer 1

use combinations

probability = (desired combinations)/ total combinations

total marbles = 24

total combinations = 24 C 6

\[P(3B) = \frac{5C3 * 19C3}{24C6}\]
\[P(0R) = \frac{15C6}{24C6}\]

Answer 2

for parts b) and d) there are separate cases which involve adding probabilities

Answer 3

So the one you did was a?

Answer 4

3B = "3 are blue"
0R = "none are red"

Answer 5

OH! Okay, thank you! I'm not good at math at all. I really appreciate it!

Answer 6

part d) its impossible for all to be Blue since there are only 5 blue marbles
so only consider All red OR All white
\[P(6R or 6W) = \frac{19C6 + 10C6}{24C6}\]

Answer 7

part b) same num of each color means there must be 2 red, 2 white and 2 blue
i was wrong about this one, its only a single case (no adding)
\[P(2R2W2B) = \frac{19C2 * 10C2 *5C2}{24C6}\]