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*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

- Thread starter MrGremlin
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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

- Feb 21, 2013

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https://www.youtube.com/watch?v=obZzOq_wSCg

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- Jan 26, 2012

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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

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- Mar 5, 2012

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Welcome to MHB, MrGremlin!

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

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:

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the deck consist of 36 cards. And 4 of them are selected and removedThere 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?

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thank you for this video, but my problem isn't solved

https://www.youtube.com/watch?v=obZzOq_wSCg

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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.

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Hearts: 6,7,8,9,10,Queen,King,AceWhich cards are initially present? I have a guess, but this should be specified in the problem statement.

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

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That is the answer to (4). Good.1)N=C^3_32=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.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

So $P(A)=\dfrac{C^7_3} {4960}$.

I'm afraid the complement rules that not work here, since "no spades" is not the complement if "only spades".P(B)= 1-(3654/4960) = 1306/4960

Right? I doubt it.

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:

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

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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?

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Erm... no.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?

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I'm afraid so.is it all wrong?

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The mistake is in what you are trying to calculate.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?

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.

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 $C^{20}_3$.P(E)-without spades and no aces

32-4-6=22

$P(E)=\dfrac {C^{20}_3} {4960} = 1140/4960$

So it should be:

$P(\text{no spades AND no aces})=\dfrac {C^{22}_3} {4960}$

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- Mar 5, 2012

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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:

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

If you combine my previous posts (see quotes), you can see that there areThis 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:

$P(\text{no spades AND no aces})=\dfrac {C^{22}_3} {4960}$

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

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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.

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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

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- Mar 5, 2012

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b) choosing the same suit and there are no face cards

You can't lump them all together.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.

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 $\clubsuit$?

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