1. Card problem

6 cards are placed face down on a table. Some are red and some are white. You guess all 6 at random.
a) Suppose you are told that 2 are red and 4 are white cards. The rule is you must guess which two are red and which two are white. What is the probability that you get k correct, k=1,2,3,4,5, or6?
b) 3 are red and 3 are white. The rule is you have to guess which 3 are red and which 3 are white. What is the probability that you get k correct, k=1,2,3,4,5,or 6?
c) 1 is red and 5 are white. What is the probability you get k correct, k=1,2,3,4,5 or 6?

How many times are you supposed to pick.... since there 2reds and 2whites you need so i thought this would mean picking 4times... reply

3. Hello, wvlilgurl!

You have a typo . . . a bit misleading.
And the instructions need some clarification.

Six cards are placed face down on a table. Some are red and some are white.
You guess all 6 at random.

a) Suppose you are told that 2 are red and 4 are white cards.
You must guess which two are red and which four are white.
What is the probability that you get k correct, k = 1,2,3,4,5,6 ?
It depends on what is meant by "guess at random."

If each guess is independent of the others, then we can use a strategy.
. . $\displaystyle P(\text{guess Red}) = \frac{1}{3},\;\;P(\text{guess White}) = \frac{2}{3}$

So it is possible that our guess might be: RRRRRW,
. . which we know is impossible,
. . but would be included in our calculations.

We could make a guess with two Reds and four Whites each time.
The probabilties can be calculated by Brute Force.

There are: .$\displaystyle {6\choose2,4} = 15$ possible ordering, listed below.

. . $\displaystyle \begin{array}{c}R\,R\,WWWW \\ R\,WR\,WWW \\ R\,WWR\,WW \\ R\,WWWR\,W \\ R\,WWWWR \end{array}$ . . . $\displaystyle \begin{array}{c} WR\,R\,WWW \\ WR,WR\,WW \\ WR\,WWR\,W \\ WR\,WWWR \\ WWR\,R\,WW \end{array}$ . . . $\displaystyle \begin{array}{c}WWR\,WR\,W \\ WWR\,WWR \\ WWWR\,R\,W \\ WWWR\,WR \\ WWWWR\,R \end{array}$

If we guess any one of the above orderings, we will guess:

. . 2 correctly $\displaystyle \frac{6}{15}$ of the time, 4 correctly $\displaystyle \frac{8}{15}$ of the time, 6 correctly $\displaystyle \frac{1}{15}$ of the time.

Therefore: . $\displaystyle \boxed{\begin{array}{ccc}P(1) &=& 0 \\ \\[-4mm] P(2) &=&\frac{2}{5} \\ \\[-4mm] P(3) &=&0 \\ \\[-4mm] P(4) &=& \frac{8}{15} \\ \\[-4mm] P(5) &=& 0 \\ \\[-4mm] p(6) &=& \frac{1}{15} \end{array}}$