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MrGremlin

New member
Mar 15, 2013
9
Three cards are randomly selected, without replacement, from a deck of 32. (without King of spades, Jack of hearts, jack of spades and queen of clubs)
Find the probability
a) That the cards are without spades and of choosing at least one ace
b) of choosing the same suit and there are no face cards
с) of choosing all suits

please help. Thanks beforehand
 

Petrus

Well-known member
Feb 21, 2013
739

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
There are 20 cards missing from the usual deck of 52, but we only know 4 of these 20. Are you certain the problem is worded correctly?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Hi MrGremlin,

Welcome to MHB! :)

I agree with MarkFL, this only makes sense if the total is 52. Then the problem becomes selecting 3 cards from 48 cards. Is that the problem you were given?

Jameson
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Three cards are randomly selected, without replacement, from a deck of 32. (without King of spades, Jack of hearts, jack of spades and queen of clubs)
Find the probability
a) That the cards are without spades and of choosing at least one ace
b) of choosing the same suit and there are no face cards
с) of choosing all suits

please help. Thanks beforehand
Welcome to MHB, MrGremlin! :)

Let's start with (a): 3 cards without spades and at least one ace.

(1) How many ways can you draw 3 cards that are no spades?
(2) How many ways can you draw 3 cards that are no spades and that contain no aces?
(3) Subtract (2) from (1) and you'll get the number of ways that there are no spades with at least 1 ace.
(4) How many ways can you draw 3 cards?
Divide (3) by (4) and you get the probability to draw 3 cards with no spades and at least 1 ace.

Btw, just like MarkFL and Jameson, it is unclear to me which cards are exactly in the deck, but these steps will work as soon as you can clarify that.
 
Last edited:

MrGremlin

New member
Mar 15, 2013
9

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Which cards are initially present? I have a guess, but this should be specified in the problem statement.
 

MrGremlin

New member
Mar 15, 2013
9
Welcome to MHB, MrGremlin! :)

Let's start with (a): 3 cards without spades and at least one ace.

(1) How many ways can you draw 3 cards that are no spades?
(2) How many ways can you draw 3 cards that are no spades and that contain no aces?
(3) Subtract (2) from (1) and you'll get the number of ways that there are no spades with at least 1 ace.
(4) How many ways can you draw 3 cards?
Divide (3) by (4) and you get the probability to draw 3 cards with no spades and at least 1 ace.

Btw, just like MarkFL and Jameson, it is unclear to me which cards are exactly in the deck, but these steps will work as soon as you can clarify that.

1)N=C^3_32=4960
P(A)- there are only spades, P(B) - there are no spades.
My deck without King of spades and Jack of spades. So I have only 7 spades.
P(A) = ((C^0_4)*(C^3_29))/4960=3654/4960
P(B)= 1-(3654/4960) = 1306/4960
Right? I doubt it.
 

MrGremlin

New member
Mar 15, 2013
9
Which cards are initially present? I have a guess, but this should be specified in the problem statement.
Hearts: 6,7,8,9,10,Queen,King,Ace
Spades:6,7,8,9,10,Queen,Ace
Diamonds:6,7,8,9,10.Jack,Queen,King,Ace
Clubs:6,7,8,9,10.Jack,King,Ace
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that was my guess...that the lowest 4 ranks (with aces high) being removed from each of the four suits.

Can you apply the method thoughtfully laid out by I like Serena?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
1)N=C^3_32=4960
That is the answer to (4). Good.

P(A)- there are only spades, P(B) - there are no spades.
My deck without King of spades and Jack of spades. So I have only 7 spades.
P(A) = ((C^0_4)*(C^3_29))/4960=3654/4960
Not sure what you did there, but with 7 spades there are $C^7_3$ ways to draw 3 cards that are only spades.
So $P(A)=\dfrac{C^7_3} {4960}$.

P(B)= 1-(3654/4960) = 1306/4960
Right? I doubt it.
I'm afraid the complement rules that not work here, since "no spades" is not the complement if "only spades".
It's also possible to have "some spades but not all spades".

However, what you can do, it to use:

$P(B) = \dfrac{\text{Number of ways that match with B}} {\text{Total number of ways}} \quad $ if each way to draw is equally likely (which it is)

Since there are 32-7=25 cards that are not spades, there are $C^{25}_3$ ways to draw 3 cards that are not spades.
So the answer to my question (1) is $C^{25}_3$.

And:

$P(B)=P(\text{no spades})=\dfrac{C^{25}_3} {4960}$
 

MrGremlin

New member
Mar 15, 2013
9
That is the answer to (4). Good.

Not sure what you did there, but with 7 spades there are $C^7_3$ ways to draw 3 cards that are only spades.
So $P(A)=\dfrac{C^7_3} {4960}$.

...

$P(B)=P(\text{no spades})=\dfrac{C^{25}_3} {4960}$

P(C) = at least one ace.= $1-\dfrac{ C^{28}_3} {N}= (4960-3276)/4960 = 1684/4960$

I try to
add the probabilities myself.

b) P(A) - without face cards. $P(A)= \dfrac{C^{20}_3} {4960} = 1140/4960$
P(B) - i don't understand. We have different number of suits. So i should find the probability for each suit
separately? Using formula P(1+2+3+4) = P(1) + P(2) ...-P(1234) -isn't correctly.

с) P(B) = different suits. P(B)= ((C^(4)_(3)* (C^(7)_(1)* (
C^(8)_(1) *(C^(9)_(1)/{4960}= 2016/4960

are these correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
P(C) = at least one ace.= $1-\dfrac{ C^{28}_3} {N}= (4960-3276)/4960 = 1684/4960$

I try to
add the probabilities myself.

b) P(A) - without face cards. $P(A)= \dfrac{C^{20}_3} {4960} = 1140/4960$
P(B) - i don't understand. We have different number of suits. So i should find the probability for each suit
separately? Using formula P(1+2+3+4) = P(1) + P(2) ...-P(1234) -isn't correctly.

с) P(B) = different suits. P(B)= ((C^(4)_(3)* (C^(7)_(1)* (
C^(8)_(1) *(C^(9)_(1)/{4960}= 2016/4960

are these correct?
Erm... no.
 

MrGremlin

New member
Mar 15, 2013
9
is it all wrong?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

MrGremlin

New member
Mar 15, 2013
9
Well. I try to explain my solution.
1) P(C) - at least one ace
there are 32 cards in my deck. 4 of them are aces. 32-4=28.
P(D)-there are no aces. $P(D)=\dfrac{C^ {28}_3} {4960}$
P(C) = 4960 - P(D)
Where is my mistake?

 

MrGremlin

New member
Mar 15, 2013
9
P(E)-without spades and no aces
32-4-6=22
$P(E)=\dfrac {C^{20}_3} {4960} = 1140/4960$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Well. I try to explain my solution.
1) P(C) - at least one ace
there are 32 cards in my deck. 4 of them are aces. 32-4=28.
P(D)-there are no aces. $P(D)=\dfrac{C^ {28}_3} {4960}$
P(C) = 4960 - P(D)
Where is my mistake?

The mistake is in what you are trying to calculate.
Problem (a) asks for no spades and at least one ace.
This is not what you are currently calculating.
You are looking at the entire deck, but you should only be looking at the cards that are no spades.

P(E)-without spades and no aces
32-4-6=22
$P(E)=\dfrac {C^{20}_3} {4960} = 1140/4960$
This looks better, you found that indeed 22 cards are available for "no spades AND no aces", but afterward you should have $C^{22}_3$ instead of $C^{20}_3$.

So it should be:
$P(\text{no spades AND no aces})=\dfrac {C^{22}_3} {4960}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Since there are 32-7=25 cards that are not spades, there are $C^{25}_3$ ways to draw 3 cards that are not spades.
So the answer to my question (1) is $C^{25}_3$.

And:

$P(B)=P(\text{no spades})=\dfrac{C^{25}_3} {4960}$
This looks better, you found that indeed 22 cards are available for "no spades AND no aces", but afterward you should have $C^{22}_3$ instead of $C^{20}_3$.

So it should be:
$P(\text{no spades AND no aces})=\dfrac {C^{22}_3} {4960}$
If you combine my previous posts (see quotes), you can see that there are

$C^{25}_3$ ways to draw 3 cards that are not spades (1).
$C^{22}_3$ ways to draw 3 cards that are not spades AND that are not aces (2).
So there are $C^{25}_3 - C^{22}_3$ ways to draw 3 cards that are not spades AND that have at least one ace (3).

So the answer to (a) is $P(\text{no spades AND at least one ace}) = \dfrac{C^{25}_3 - C^{22}_3}{4960}$.
 

MrGremlin

New member
Mar 15, 2013
9
I like Serena, Thank you very much.:)

So, b) P(A) - no face cards. there are 12 face cards in the deck.(including aces)
30-12 = 20 cards.
$P(A)= \dfrac {C^{20}_3} {4960}$
P(B)= the same number of colors. $P(\clubsuit + \diamondsuit+ \heartsuit+ \spadesuit)= 1 - P(\overline{\clubsuit + \diamondsuit+ \heartsuit+\spadesuit})$, but they are incompatible events. So $P(\overline{\clubsuit + \diamondsuit+ \heartsuit+\spadesuit})=0$ and P(B) = 1.







- - - Updated - - -

c)The number of cards by suits 7, 8, 8, 9. i.e. the number of successful samples = 7*8*8 +7*8*9 +7*8*9+8*8*9= 2032
P(A)= 2032/4960
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
b) choosing the same suit and there are no face cards
I like Serena, Thank you very much.:)

So, b) P(A) - no face cards. there are 12 face cards in the deck.(including aces)
30-12 = 20 cards.
$P(A)= \dfrac {C^{20}_3} {4960}$
P(B)= the same number of colors. $P(\clubsuit + \diamondsuit+ \heartsuit+ \spadesuit)= 1 - P(\overline{\clubsuit + \diamondsuit+ \heartsuit+\spadesuit})$, but they are incompatible events. So $P(\overline{\clubsuit + \diamondsuit+ \heartsuit+\spadesuit})=0$ and P(B) = 1.
You can't lump them all together.
It matters which suit you pick.

So let's start with the first suit $\clubsuit$.
How many cards that are not facecards?
Yielding how many ways to draw 3 cards in the single suit $\clubsuit$?

$P(\text{only }\clubsuit \wedge \text{no face cards in }\clubsuit) =?$