SOLUTION: (a) Let X be a random variable giving the number of aces in a random draw of 4 cards from an ordinary deck 52 cards. Construct a table showing the probability distribution of X.

Question 1193236: (a) Let X be a random variable giving the number of aces in a random draw of 4 cards from an ordinary deck 52 cards. Construct a table showing the probability distribution of X.
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

There are four aces in a deck. One of each suit (hearts, diamonds, clubs, spades).

X = number of aces in a four card hand

The allowed values of X are drawn from the set {0,1,2,3,4}

If X = 0, then we have 52-4 = 48 non-ace cards to pick from.
We have four slots to fill. Order does not matter in card hands, meaning we use the nCr combination formula.
Plug in n = 48 and r = 4
n C r = (n!)/(r!(n-r)!)
48 C 4 = (48!)/(4!*(48-4)!)
48 C 4 = (48!)/(4!*44!)
48 C 4 = (48*47*46*45*44!)/(4!*44!)
48 C 4 = (48*47*46*45)/(4!)
48 C 4 = (48*47*46*45)/(4*3*2*1)
48 C 4 = (4669920)/(24)
48 C 4 = 194580
There are 194,580 ways to select four non-ace cards

Next, calculate the number of ways to select four cards regardless if we get an ace or not.
n = 52 and r = 4
n C r = (n!)/(r!(n-r)!)
52 C 4 = (52!)/(4!*(52-4)!)
52 C 4 = (52!)/(4!*48!)
52 C 4 = (52*51*50*49*48!)/(4!*48!)
52 C 4 = (52*51*50*49)/(4!)
52 C 4 = (52*51*50*49)/(4*3*2*1)
52 C 4 = (6497400)/(24)
52 C 4 = 270725
We have 270,725 ways to select any four cards whether we get an ace or not.

I should mention that the cards are NOT put back. We consider this scenario as "no replacement".

Once we have the two results calculated earlier, we divide them
194580/270725 = 0.718737

The probability that we get zero aces, i.e. X = 0, is roughly 0.718737

In the table below, we'll have P(X) = 0.718737 pair up with X = 0 in the first row.
There's roughly a 71.8737% chance of getting no aces in a four-card hand.

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Now onto X = 1 which means we have exactly one ace in our four-card hand.

We have four aces, so there are four ways to select exactly one of them.
You can use the nCr formula to note that 4C1 = 4, though this may be a bit overkill.

For the remaining r = 4-1 = 3 slots, we have n = 52-4 = 48 cards to pick from
n C r = (n!)/(r!(n-r)!)
48 C 3 = (48!)/(3!*(48-3)!)
48 C 3 = (48!)/(3!*45!)
48 C 3 = (48*47*46*45!)/(3!*45!)
48 C 3 = (48*47*46)/(3!)
48 C 3 = (48*47*46)/(3*2*1)
48 C 3 = (103776)/(6)
48 C 3 = 17296
which is the number of ways to pick the other 3 non-ace cards.

We found there are 4 ways to pick the single ace, and 17296 ways to pick the other three cards.
This then means there are 4*17296 = 69184 ways to get a four-card hand such that we have exactly one ace.

Divide this over the 270725 figure found in the previous section (it represents the number of four-card hands total)
69184/270725 = 0.255551
This is the approximate P(X) value for X = 1
Refer to the second row of the table below.

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Next up is X = 2
We have 4C2 = 6 ways to select the two aces and 48C2 = 1128 ways to pick the other two non-ace cards.
I'm skipping steps, but you'd use the nCr formula as described in the other sections above (with different n and r values of course).

So we have (4C2)*(48C2) = (6)*(1128) = 6768 ways to select exactly two aces and two non-aces.
This leads to 6768/270725 = 0.025000 as the approximate P(X) value when X = 2
The more accurate P(X) value is P(X) = 0.02499953827684 but I'm rounding results to six decimal places.
Feel free to use different decimal precision. Of course, if your teacher instructs otherwise, then use that precision.

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For X = 3, we have 4C3 = 4 ways to pick the three aces.
This is the same as having 4 ways to not pick a particular ace.
Then we have 48C1 = 48 ways to pick the non-ace.

There are (4C3)*(48C1) = (4)*(48) = 192 ways to pick exactly three aces and some other card.
192/270725 = 0.000709 approximately
Therefore, if X = 3 then P(X) = 0.000709 roughly

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Lastly, we need to find P(X) when X = 4
There's only one way to pick four aces. You could use the fact that 4C4 = 1.

This is out of 270725 ways to select a four-card hand
1/270725 = 0.000004 approximately
The more accurate value is 0.0000036937852 but I'm rounding to 6 decimal places like mentioned earlier.

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You probably were able to spot the pattern how to calculate any general P(X) value fairly quickly.

That pattern would be:
P(X) = ( (4 C w)*(48 C v) )/(52 C 4)
where w+v = 4 and w,v are integers from the set {0,1,2,3,4}
w = number of aces, v = number of non-aces

For example, we could have w = 0 and v = 4
P(X) = ( (4 C w)*(48 C v) )/(52 C 4)
P(X) = ( (4 C 0)*(48 C 4) )/(52 C 4)
P(X) = ( (1)*(194,580) )/(270,725)
P(X) = 0.71873672545941
P(X) = 0.718737
which matches up with the P(X) value we found when X = 0

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Answer:

X	P(X)
0	0.718737
1	0.255551
2	0.025000
3	0.000709
4	0.000004

where X is the number of aces of a four-card hand

The P(X) values should add up to 1, or very close to it.
In this case,
0.718737+0.255551+0.025000+0.000709+0.000004 = 1.000001
which unfortunately is due to rounding error; however, we're close enough.

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