# Thread: Probability for playing cards

1. ## Probability for playing cards

Mr.A has 13 cards of the same suit. He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card, i.e he will take 3 from king whose value is 13, 0 from 10, 9 from 9 and so on.. What is the probability that he can form a number that is divisible by 2?

My working
(which is incorrect):

Since this question is dealing with combination. I could answer it as:

6C4/13C4 .. (6 possible ways to select numbers ending with an even digit with 4 chosen cards)

The correct answer according to the book is: (6 * 12 * 11 * 10)/(13 * 12* 11 * 10) = 6/13

The solution from the book went above my head. How do i solve this question?

2. Hello, saberteeth!

Mr.A has 13 cards of the same suit.
He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card,
That is: .Ace = 1, Deuce = 2, Trey = 3, . . . Ten = 0, Jack = 1, Queen = 2, King = 3.
What is the probability that he can form a number that is divisible by 2?
"Divisible by 2" means that the 4-digit number is even.

There are: 7 odd digits and 6 even digits.

He can make an even number if at least one digit is even.

He will fail of all four digits are odd.
. . There are: .$\displaystyle {7\choose4} = 35$ ways to get 4 odd digits.
. . . . There are: .$\displaystyle {13\choose4} = 715$ possible outcomes.
. . Hence: .$\displaystyle P(\text{odd number}) \:=\:\frac{35}{715} \:=\:\frac{7}{143}$

Therefore: .$\displaystyle P(\text{even number}) \;=\;1-\frac{7}{143} \;=\;\frac{136}{143}$

3. Hello,

Soroban's way is the way I would do it but it also easy to see it this way. Being divisible by 2 means it has to be even so out of the 13 cards, there are 6 even cards {2,4,6,8,10,12} . The probability of the last number being even is 6 out of 13 cards. 6/13 !

*another way of looking at it same concept:

I don't know if this is notationally correct but:

Sample Space S = {1,2,3,4,5,6,7,8,9,10,11,12,13}
Even ={2,4,6,8,10,12}

6 of the 13 cards are even; each has the same probability of being chosen. So P(EVEN)=P(2)+P(4)+...+P(12)= (1/13+1/13+1/13+1/13+1/13+1/13)= 6/13.

This way is trivial though and is not very reliable (not to mention long) when you start getting into more complex problems.

4. Again, it can be simplified by calculating the probability they'd all be odd

$\displaystyle P_{Even}=1-P_{Odd}=1-\left(\frac{7}{13}\ \frac{6}{12}\ \frac{5}{11}\ \frac{4}{10}\right)$

$\displaystyle =1-\frac{7(6)5(4)}{13(12)11(10)}=1-\frac{7(6)}{13(11)6}=1-\frac{7}{143}$

5. Originally Posted by Soroban
Hello, saberteeth!

"Divisible by 2" means that the 4-digit number is even.

There are: 7 odd digits and 6 even digits.

He can make an even number if at least one digit is even.

He will fail of all four digits are odd.
. . There are: .$\displaystyle {7\choose4} = 35$ ways to get 4 odd digits.
. . . . There are: .$\displaystyle {13\choose4} = 715$ possible outcomes.
. . Hence: .$\displaystyle P(\text{odd number}) \:=\:\frac{35}{715} \:=\:\frac{7}{143}$

Therefore: .$\displaystyle P(\text{even number}) \;=\;1-\frac{7}{143} \;=\;\frac{136}{143}$

Hello:

I did not understand one thing in Soroban's answer:

"He can make an even number if at least one digit is even."

How about $\displaystyle 2343,1225,8643$, they also contain at least one even digit but still odd.
I think it should be that last digit should be an even number i.e $\displaystyle 3334,7134$ etc.

6. Hi u2_wa,

In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.

The book answer gives $\displaystyle \frac{6}{13}$

which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.

However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.

7. Originally Posted by Archie Meade
Hi u2_wa,

In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.

The book answer gives $\displaystyle \frac{6}{13}$

which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.

However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.

What I did to solve is this:

There are in total $\displaystyle 13P4$ ways.

There should $\displaystyle (2,4,6,8,0,2)$ be among $\displaystyle six$ digits in the end to make an even number.

Ways to get an even number: $\displaystyle 12P3*6$

probability$\displaystyle =\frac{12P3*6}{13P4}$

$\displaystyle =\frac{6}{13}$ (The answer given in the book)

8. Yes, that's good work u2_wa,

shows you can work it out in alternative ways.

You are calculating the probability if we must take the digits
in the order they come in.

The way the question is worded suggests that Mr. A chooses the 4 cards and tries to
make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.

Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.

However, the way the question is worded suggests he is!!

Hence, wording can make quite a difference in probability questions.

9. Originally Posted by Archie Meade
Yes, that's good work u2_wa,

shows you can work it out in alternative ways.

You are calculating the probability if we must take the digits
in the order they come in.

The way the question is worded suggests that Mr. A chooses the 4 cards and tries to
make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.

Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.

However, the way the question is worded suggests he is!!

Hence, wording can make quite a difference in probability questions.
Yeh I know, thanks!!!