Probability

The probability of all three getting a red card is:

 

[3 / 6]  x  [2 / 5]  x  [1 / 4] ==6 / 120 ==1 / 20

 

The answer is 2/5

 

Think of the cards as all being different 3 different red and 3 different blue

There are 6!  ways that these can be ordered.

 

now I am going to give the first person 2 cards then the second and then the 3rd.

 

6 choices to start

then 3 of the ones left are the other colour

 

now there are 4 cards the second person can get any of them

there are 2 that are a different colour for the seconds person second choice

 

there are 2 cards left for the 3rd peron to chose from and then just 1

so that is

 

6*3*4*2*2

 

so the prob will be  (6*3*4*2*2)/ 6!  =  2/5

Each person gets 2 cards.

If we consider the cards to be distinguishable, there are 3! ways in which the 3 red cards can be distributed, 1 each, to the 3 persons. There are similarly 3! ways in which the remaining cards can be distributed. Thus, there are 3!×3! ways to give every person a red card.

There are 6! undiscriminated ways to distribute 6 cards.

Therefore, the probability that each person gets 1 red card is:

[3!×3!] / 6!=1 / 20.

Adjudicating.

Label the cards \(\displaystyle R_{1},R_{2},R_{3},B_{1},B_{2},B_{3}.\)

Give person number 1 two random cards, 6C2 ways of doing that.

Give person number 2 two of the remaining four cards, 4C2 ways of doing that.

Person number 3 gets what's left.

The total number of ways of doing that is 6C2*4C2 = 15*6 = 90.

Now person number 1 is to have 1 red card and 1 blue card. There are 3*3 = 9 ways in which that can happen.

Person number 2 is to have 1 red card and 1 blue card from the remaining 4 cards. That can happen in 2*2 = 4 ways.

Person number 3 has what's left.

So, the number of ways in which each person has 1 red card and 1 blue card is 9*4 = 36.

So, the probability of each person has 1 red card and 1 blue card is 36/90 = 2/5.