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3) In another game, there are 10 marbles in a box. 6 are red and 4 are blue. You
pull one marble from the box, don't put it back, and then pull another marble.
Calculate the following probabilities:
a) Pulling two reds.
b) Pulling two blues.
c) Pulling one of each.

Sagot :

Part (A)

We have 6 red and 4 blue marbles, so 6+4 = 10 total.

The probability of getting red is 6/10 since we have 6 red out of 10 total.

The probability of getting another red is 5/9 because we don't put the first marble back.

The probability of getting two reds in a row is (6/10)*(5/9) = 30/90 = 1/3

Answer: 1/3

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Part (B)

4/10 represents the probability of getting blue on the first try

3/9 represents the probability of getting a blue marble on the second try

We don't put the marble back.

(4/10)*(3/9) = 12/90 = 4/30 is the probability of getting two blues in a row.

Answer: 4/30

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Part (C)

Add up the results of parts (A) and (B)

1/3 + 4/30

10/30 + 4/30

(10+4)/30

14/30

7/15

The result 7/15 is the probability of either getting two reds OR getting two blues (pick one case only). This tells us that 7 out of 15 times we'll have this happen. This must mean the remaining 8 times will have two marbles of different colors.

So 8/15 is the probability of picking one marble of each color.

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As an alternative, we could also consider it like this

(6/10)*(4/9) = 24/90 = 4/15

This represents the probability of picking red first, then blue next.

Similarly,

(4/10)*(6/9) = 24/90 = 4/15

represents the probability of picking blue first, then red next.

Therefore, 4/15+4/15 = 8/15 is the probability of getting one of each color.

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Answer:   8/15

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