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A probablity question...(its easy but i am not able 2 solve)

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by mathurmudit1003 » Wed Jun 20, 2012 1:08 pm
In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of
the bag, what is the probability that he will have one white and one blue marble?

need explanation for the solution
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Source: — Problem Solving |

by eagleeye » Wed Jun 20, 2012 3:02 pm
Hi mathurmudit1003:

There are 2 ways in which you can easily do it:
1. Selections:

Here we have 3R, 2W, 5R for a total of 10 marbles. We need to find the probability of selecting one white and one blue.

a. No. of ways of selecting 2 marbles out of 10 = 10C2 = 45
b. No. of ways of selecting 1 white marble out of 2 = 2C1 = 2
c. No. of ways of selecting 1 red marble out of 5 5C1 = 5

Now the probability of selecting one white AND one red = (2C1*5C1)//(10C2) = 2*5/45 = 2/9.

2. Arrangements:
We need to select two marbles - one white and one blue. There are two arrangements (order is important) that you can do with those.

a. Select white one first, select red one second.
b. Select red one first, select white one second.

Now probability of selecting the first one white = (No. of whites)/(Total) = 2/10
Probability of selecting the second one red (after selecting the first one) = (No. of Reds)/(Remaining Total) = 5/9.

So probability of selecting first white AND second red = 2/10*5/9.
Same way probability of selecting red one first AND white one second = 5/10*2/9.

So probability required = SUM of the two cases above = 2/10*5/9 + 5/10*2/9 = 20/90 = 2/9.
Let me know if this helps :)
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