There are 2 blue and 3 red balls in a bag. If two balls are

Question

There are 2 blue and 3 red balls in a bag. If two balls are picked w/o replacement, what is the prob of getting at least 1 blue?

I tried it this way and messed up:

   All outcomes   
10C5 = 10

   Fav outcomes   
2C1 = 2

2C2 = 2(1)/(2)(0) (there's obviously something wrong here)

Can anyone explained where I messed up and whether this method is applicable? I've since realized it's not the quickest way but I need to know why this didn't work.

pau.sabria · Answer

2C2= 2!/2!0!

What I assume you did not new is that 0!=1

Anyways... what I would do is the following:

the probability of having at least one blue is the complementary probability of getting 2 red balls withouth replacement. That is:

P(at least 1 blue ball)=1 - P(two red balls)

P(two red balls)=P(1st ball=red)Â·P(2nd ball=red)

Since there is no replacement:

P(1st ball=red)= 3/5
P(2nd ball=red)=2/4 (2 red balls left out of 4 balls)

P(at least 1 blue ball)= 1-3/5Â·2/4 = 1-6/20 = 14/20=7/10=70%

Hope it helps!

ggarr · Answer

Yes. Your solution is the quicker solution I soon realized. Thanks very much. But can anyone tell me what is wrong w/the combo method?

All outcomes 
10C5 = 10 

Fav outcomes 
2C1 = 2 

2C2 = 2(1)/(2)(0!) = 1

2+1/10 = 3/10 - what am I missing?

devilmirror · Answer

Umm, 2C2 = (2!)/(2! x 0!) 
and 0! = 1

Therefore, 2C2 = 2/(2 x 1) = 1

devilmirror · Answer

Now I will solve this question by using your approach. Your method is correct but need a little bit of adjustment in logic. :wink:

Question: Find Prob This is kind of advance probability question. You can split Prob into 2 cases. Prob = Prob + Prob Prob = (pick 1 blue) x (pick 1 red) / (pick 2 ball randomly) = (2C1) x (3C1)/(5C2) = (2 x 3) / (10) = 6/10 Prob = (pick 2 blue) / (pick 2 ball randomly) = (2C2) / (5C2) = 1/10 Therefore; Prob = 6/10 + 1/10 = 7/10 Or, you can use another method,which has already been shown by one of our friend above me, to solve this question. Prob = 1 - Prob Prob = (3C2)/(5C2) = 3/10 Prob = 1 - Prob = 1 - 3/10 = 7/10

amorpheus · Answer

I will solve it like this:

1 blue ball: 2C1 * 3C1 / 5C2 = 6/10 = 0.6
both balls are blue: 2C2/10 = 1/10 = 0.1

total p = 0.6 + 0.1 = 0.7

There are 2 blue and 3 red balls in a bag. If two balls are

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