1. ## Simple Probability Question

I am a bit confused by the way this question is solved:-

From a of well shuffled pack 52 cards, three cards are drawn at random. Find the probability of drawing an ace, a king and a jack.

Solution given:- There are 4 aces, 4 king and 4 jacks and their selection can be made in the following ways:
12C1 X 8C1 X 4C1 = 12 X 8 X 4.
Total selections can be made = 52C3= 52 X 51 X 50.
Therefore required probability = $\displaystyle \frac{(12)(8)(4)}{ (52)(51)(50)}$

I don't understand why are we taking 12C1 X 8C1 X 4C1 = 12 X 8 X 4 instead of 4C1 X 4C1 X 4C1 = 4 x 4 X 4 for the numerator. Since, we are selecting 1 ace from 4 aces, 1 king from 4 kings and 1 jack from 4 jacks shouldn't we be taking 4C1 X 4C1 X 4C1 = 4 x 4 X 4 for the favourable events ? Please advice on the above.

2. ## Re: Simple Probability Question

Originally Posted by SheekhKebab
I am a bit confused by the way this question is solved:-

From a of well shuffled pack 52 cards, three cards are drawn at random. Find the probability of drawing an ace, a king and a jack.
Here's how I would do that: there are 52 cards, 4 aces, four kings, and a jack. The probability the first card drawn is an ace is 4/52= 1/13. There are then 51 cards left, flur of which are kings. The probability that the second card you draw is a king is 4/51. There are then 50 cards left, 4 of which are jacks. The probability the third card you draw is a jack is 4/50= 2/25. The probability of drawing "ace, king, jack" in that order is (1/13)(4/51)(2/25)= 8/(13*51*25). But if you look at "jack, ace, king" or any other specific order you will see that while you have different fractions, you have the same numerators and the same denominators in different orders so the same probability. There are 3!= 6 such orders so there the probability of drawing an ace, king, and jack is 6(8/(13*51*25)).

Solution given:- There are 4 aces, 4 king and 4 jacks and their selection can be made in the following ways:
12C1 X 8C1 X 4C1 = 12 X 8 X 4.
Total selections can be made = 52C3= 52 X 51 X 50.
Therefore required probability = $\displaystyle \frac{(12)(8)(4)}{ (52)(51)(50)}$

I don't understand why are we taking 12C1 X 8C1 X 4C1 = 12 X 8 X 4 instead of 4C1 X 4C1 X 4C1 = 4 x 4 X 4 for the numerator. Since, we are selecting 1 ace from 4 aces, 1 king from 4 kings and 1 jack from 4 jacks shouldn't we be taking 4C1 X 4C1 X 4C1 = 4 x 4 X 4 for the favourable events ? Please advice on the above.

The reason for "$\displaystyle ^{12}C_1$" is that a total of 12 "aces, kings, and jacks" and you are drawing one of them. The reason for the "$\displaystyle ^8C_1$" is that whether that is an ace, king, or jack, there are then 8 of the remaining kind of card you are looking for (if the first card drawn was a jack, there are 8 aces and kings left) and you want to draw one of them. The reason for the "$\displaystyle ^4C_1$" is that whichever of ace, jack, or king, the first two cards are, there are 4 cards of the remaining type and you want to draw 1 of them.

3. ## Re: Simple Probability Question

Hi HallsofIvy,

Thanks ! That 3! I would have definitely factored in if I had solved the question completely. Since the 3! would have come from the denominator of 52C3. So, that was not my question. I was a bit confused about the way the selection was made in the numerator of the original solution.

So, according to you, then, both the solutions/approach are correct ? Should we prefer one approach over the other cause the original solution doesn't appear to be much convincing ?

4. ## Re: Simple Probability Question

Both solutions are correct, and both give the same answer. Neither is "more correct" than the other, though I would admit that my instinct is to solve it the way that Halls did. I tend to think first in terms of selecting cards in a particular order, then multiply by the number of ways that the order can be changed, which is what he did. But the book's solution is equally valid, and arguably more elegant - i.e. how many choices do I have for the first card, then how many choices are available for the 2nd card, then how many for the third card, with no need to go through an additional step of considering the specific order of cards selected.

5. ## Re: Simple Probability Question

Hi Ebaines,

Thanks ! But for the original solution I think there is an error in the denominator . 52C3=$\displaystyle \frac{(52)(51)(50)}{3!}$, but that 3! is missing in the original solution. So we are not getting an answer of $\displaystyle \frac{16}{5525}$ which is the correct answer and which we will get if we follow the other approach. Please inform whether I am correct or I am missing something !

6. ## Re: Simple Probability Question

Both approaches yield the same results: the first is

$\displaystyle \frac {12 \times 8 \times 4}{52 \times 51 \times 50} = 0.002896$

and HallsOfIvy's approach:

$\displaystyle \frac {6 \times 8}{13 \times 51 \times 25} = 0.002896$

Note that you can multiply the numeratort and denominator of Halls' by 8 get the same form as the first:

$\displaystyle \frac {6 \times 8}{13 \times 51 \times 25} \times \frac 8 8 = \frac {12 \times 8 \times 4}{52 \times 51 \times 50} = 0.002896$

The first approach does not need to explicitly multiply by 3! because the fact that the three cards may be selected in any order is already included in the numerator by using 12 x 8 x 4.

7. ## Re: Simple Probability Question

Hi ebaines,

But the original solution mentions: Total selections can be made in 52C3 ways, which is equivalent to $\displaystyle \frac{(52)(51)(50)}{3!}$, which is the denominator. So where will the 3! go then ?

8. ## Re: Simple Probability Question

This part of your first post is incorrect:

"Total selections can be made = 52C3= 52 X 51 X 50."