probability

MrGremlin

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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.
 
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.

Are you sure that you have deck 32 - and you have removed only 4 cards?

Also please tell us how does your book/classnotes explain probability?

Please share your opinions and your reasons behind those opinions - with us.

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1)My deck consists of 36 cards.
(
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)
And 4 of them are selected and removed

I doubt in my solution, but:

N= С^3_32=4960
(without
the addition of probabilities)
а) P(A)- there are no spades P(A) =
{C^{25}_3} /{4960}/N = 2300/4960
P(B) -at least one ace.= 1-(C^28_3)/N= (4960-3276)/4960 = 1684/4960
is it right?


 
1)My deck consists of 36 cards.36?
(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)
And 4 of them are selected and removed

I doubt in my solution, but:

N= С^3_32=4960
(without
the addition of probabilities)
а) P(A)- there are no spades P(A) =
{C^{25}_3} /{4960}/N = 2300/4960 What does 3 have to do with this problem? You are drawing 4 cards.
P(B) -at least one ace.= 1-(C^28_3)/N= (4960-3276)/4960 = 1684/4960
is it right?
You say you have 8 hearts, 7 spades, 9 diamonds, and 8 clubs, which add up to 32. You also say you have 36 cards. Which is it?

Assuming it is 32

Problem a

Number of ways to select 4 cards from 32 in deck = \(\displaystyle \dbinom{32}{4} = \dfrac{32!}{4! * 28!} = \dfrac{32 * 31 * 30 * 29}{4 * 3 * 2 * 1} = 4 * 31 * 10 * 29 = 35,960.\)

Number of ways to select 4 cards from (32 - 7) non-spades in deck =

\(\displaystyle \dbinom{32 - 7}{4} = \dfrac{25!}{4! * (25 - 4)!} = \dfrac{25 * 24* 23* 21}{4 * 3 * 2 * 1} = 25 * 3 * 23 * 7 = 12,075.\)

Probability of no spades in 4 draws = \(\displaystyle \dfrac{12,075}{35,960} \approx 33.58\%.\)

I have no idea what you are doing on problem b. At least one ace means one ace, or two aces, or three aces, or four aces. So one way to solve is to calculate the probabilities of each one of those mutually exclusive events and add them up. Alternatively, you can recognize that at least one means not zero, calculate the probability of zero aces and then do what?
 
King of spades, Jack of hearts, jack of spades and queen of clubs - these cards are selected and removed. I need to find the probability of drawing 3 cards
 
What you wrote was very confusing! You first said just that three cards were selected from 32 then wrote that you had a deck of 36 from with 4 were removed (That's a very peculiar deck. I know that a pinochle deck has only 48 cards but I have never heard of one with only 36!) You then say
"My deck consists of 36 cards.
(
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)

And 4 of them are selected and removed"

which makes no sense because you have given, here 32 cards so apparently your deck consists of 6 and higher in each suit with the four cards already remove NOT with "4 of them selected and removed".
 
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