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Question

You have 3 baskets.

Each basket contains 4 balls, indistinguishable except for color.

There is a red, a white, a blue, and a black ball in each basket.

You are blindfolded and asked to remove one ball from each basket.

What are the chances you will pick exactly 2 red balls and 1 non-red ball?

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The long way is...

R = Red, W = White, B = Blue, K = Black

RRR

rrw

rrb

rrk

rwr

RWW

RWB

RWK

rbr

RBW

RBB

RBK

rkr

RKW

RKB

RKK

wrr

WRW

WRB

WRK

WWR

WWW

WWB

WWK

WBR

WBW

WBB

WBK

WKR

WKW

WKB

WKK

brr

BRW

BRB

BRK

BWR

BWW

BWB

BWK

BBR

BBW

BBB

BBK

BKR

BKW

BKB

BKK

krr

KRW

KRB

KRK

KWR

KWW

KWB

KWK

KBR

KBW

KBB

KBK

KKR

KKW

KKB

KKK

The lower case letters meet the criteria. 9/64

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now let's change it...

You are now asked to remove two balls from each basket (ie 6 balls total). Are you more likely to pick exactly [1 red ball and 5 non-red balls] or exactly [2 red balls and 4 non-red balls]? What are the chances? :wacko:

extra credit - anyone can run the options out the long way (I did B)) ) - but can you explain the math behind the result? I left my high-school math way behind, so don't have a clue what the formula would look like...

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Are you more likely to pick exactly [1 red ball and 5 non-red balls] or exactly [2 red balls and 4 non-red balls]?

What are the chances?

No.

You're not more likely to pick 1 red ball or 2 red balls.

The chances are 3/8 for both outcomes.

Basically, if you pick 2 balls from a basket, half the time you'll get the red one [RW, RB, RK], and half the time you won't [WB, WK, BK].

Of the 8 outcomes of picking from three baskets, there are 3 ways to get 1 red [100 010 001], and 3 ways to get 2 red [110 101 011].

There are also 1/8 chances for 0 red [000] and 3 red [111].

Going a little slower,

With regard to getting red balls when picking from all three baskets, there are 8 equally likely outcomes.

You get 1 red ball in three of the outcomes and 2 red balls in three other outcomes:

0 0 0 = 0 red balls

1 0 0 = 1 red balls

0 1 0 = 1 red balls

0 0 1 = 1 red balls

1 1 0 = 2 red balls

1 0 1 = 2 red balls

0 1 1 = 2 red balls

1 1 1 = 3 red balls

Probability of getting 0 red balls taking 2 from each basket = 1/8

Probability of getting 1 red balls taking 2 from each basket = 3/8

Probability of getting 2 red balls taking 2 from each basket = 3/8

Probability of getting 3 red balls taking 2 from each basket = 1/8

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