A jar contains a mixture of 12 black marbles, 10 red marbles, and 18 white marbles, all the same size. If two marbles are drawn from the jar without being replaced, what would the probability be: 1. of drawing a white, then a black marble? 2. of drawing two black marbles? 3. of drawing a black, then a red marble? 4. of drawing two white marbles?

Question

Mathematics

Cheyenne Newton

20 January, 12:41

A jar contains a mixture of 12 black marbles, 10 red marbles, and 18 white marbles, all the same size. If two marbles are drawn from the jar without being replaced, what would the probability be: 1. of drawing a white, then a black marble? 2. of drawing two black marbles? 3. of drawing a black, then a red marble? 4. of drawing two white marbles?

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Brayan Castillo · Answer 1

Sample space = {12 B; 10 R; 18 W} = 40 = All possibilities

1) P (1 W) = Favorable Space/All possibility = 18/40 = 0.45

No replacement: 2nd draw P (1 B) = 12 / (39) = 0.307 (39 because 1 W already drawn & not replaced)

Now P (1 W AND 1 B) = P (1 W ∩ 1 B) = 0.45 x 0.307 = 0.138

2) P (1 B) = 12/40 = 0.25, and another B; P (11/39) = 0.282 (NO replacement)

P (1 B and 1 B) = 0.25 x 0.282 = 0.07

3) P (1 B) = 12/40 = 0.25

P (1 R) = 10/39 = 0.256

P (1 B AND 1 R) = 0.25 x 0.256 = 0.064

4) P (1 W) = 18/40 = 0.45

P (of another W) = P (17/39) = 0.435

P (1 W AND 1 W) = 0.45 x 0.435 = 0.195

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