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The King and Four Queens (1956)
IMDB rating: 5.70
Plot: Smooth cowboy Dan Kehoe arrives at a ranch run by an old widow and her four daughters-in-law. He’s been tipped off that the proceeds of a gold robbery are hidden on the ranch and only one of the women knows where. He plays them off against each other in his quest to discover the location.
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The King and Four Queens
Directors: Walsh Raoul
Actors: Gable Clark,Roberts Roy,Shields Arthur,Flippen Jay C.,Ames Florenz,Roberson Chuck,Adventure,Comedy,Mystery,Western,
Help with probability…..?
Help with probability…..deck of cards?
helppp i need some help!
explain, show me how to do this , what to do?
suppose u have a reduced deck of cards that has only the 10s ,jacks ,queens, kings, and aces of the four suits, for total of 20 cards.
u draw one card replace it and draw another card
P(10 and ace)
P(queen and king)
P(two hearts)
P(two clubs)
you draw one card and replace it then draw another and replace it and then draw another card and replace it
P(three of a kind, any card)
P(three jacks)
you draw four cards and replace each card after drawing
P(four kings)
P(four of any kind,any card)
same deck of cards though without replacement
u draw two cards at once without replacement
P(10 and ace)
P(queen and king)
P(two hearts)
P(two clubs)
this time three cards
P(three of any kind,any card)
P(three jacks)
now four
P(four kings)
plz and thanks
sorry if it is too much
i guess i didnt realize it when typing the problem out
ok….i sort of got with replacement..
though i kind of dont get the second half
which is all the without replacement
and could u leave the answer in fraction form?
thanks
WITH REPLACEMENT:
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P(10 and A) with replacement:
This could mean the probability of drawing a 10 on the first draw, then replacing it and drawing an Ace on the second draw. Or, it could mean the probability of drawing a 10 and Ace in either order.
In the first case (order matters):
Draw a 10 with probability 1/5, then draw an Ace with probability 1/5, with a likelihood:
P(10 then A) = (1/5)*(1/5) = 1/25
In the second case (order doesn’t matter):
There are 20^2 = 400 ways to draw two cards with replacement.
There are 8 ways to draw either a 10 or Ace on the first draw, and 4 ways to draw the corresponding Ace or 10 on the second draw.
P(10 and A) = (8)(4) / 400 = 2/25
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P(Q and K): This is exactly the same as the P(10 and A) problem.
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P(two hearts) with replacement:
Draw a heart with probability 1/4, replace the card and draw another heart with probability 1/4:
P(two hearts) = (1/4)(1/4) = 1/16
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P(two clubs): This is exactly the same as two hearts.
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P(three of a kind) with replacement:
Select any rank on the first draw. Replace it and draw a card of the same rank with probability 1/5. Replace it, and draw another card of the same rank with probability 1/5:
P(three of a kind) = (1/5)(1/5) = 1/25
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P(J, J, J) with replacement:
Draw a jack on the first, second and third draws each with probability 1/5:
P(J, J, J) = (1/5)^3 = 1/125
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P(K, K, K, K) with replacement:
Draw a king on all four draws each with probability 1/5:
P(K, K, K, K) = (1/5)^4 = 1/625
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P(four of a kind) with replacement:
Draw a card of any rank and replace it. Then, on the following three draws, draw cards matching the first card’s rank, each with probability 1/5:
P(four of a kind) = (1/5)^3 = 1/125
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WITHOUT REPLACEMENT:
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P(10 and A) without replacement
There are 20C2 ways to choose 2 cards out of 20
Each of the four 10s may be paired with each of the four Aces in 4*4 = 16 ways.
P(10, A) = 16 / 20C2 = 16/190 = 8/95
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P(Q and K): This is exactly the same as P(10, A)
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P(two hearts) without replacement
There are 5C2 ways to choose two hearts, out of 20C2 total two-card combinations:
P(two hearts) = 5C2 / 20C2 = 10/190 = 1/19
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P(two clubs): This is the same as two hearts.
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P(three of a kind) without replacement
There are 20C3 ways to choose 3 cards.
To make three of a kind, in in any 5 ranks, choose three cards in a rank in 4C3 ways.
P(three of a kind) = (5) (4C3) / 20C3 = 20/1140 = 1/57
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P(J, J, J) without replacement
Choose any three out of the four Jacks in 4C3 ways:
P(J, J, J) = 4C3 / 20C3 = 4/1140 = 1/285
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P(K, K, K, K) without replacement.
Out of 20C4 ways to draw four cards, there’s only one way to draw four Kings.
P(K, K, K, K) = 1 / 20C4 = 1/4845
Hope that helps!
rangergordon | Jan 09, 2010
Well this is a lot of questions for just 10 points. Fortunately they are all done the same way.
Since you are replacing the card after each draw the problem is much, much simpler.
Pr(10 and ace) = Pr(10) Pr(Ace) + Pr(Ace)Pr(10) = 2(4/20) (4/20) = 32/400 = 2/25
That is–the probability you get a 10 and an ace is equal to the probability you draw a 10, and then an ace. you also need to account for the probability you draw an ace, and then a 10.
Pr(queen and king) = Pr(queen) Pr(king) + Pr(king)Pr(queen) = 2(4/20) (4/20) = 16/400 = 2/25
Pr(two hearts) = Pr(Heart) Pr(Heart) = (5/20)(5/20) = 1/16
For Pr(two hearts) there is only one way to draw two hearts, so you don’t have to multiply by two like we did above.
Pr(two clubs) = Pr(Club)Pr(Club) = (5/20)(5/20) = 1/16
P(three of a kind, any card) = Pr(Three 10s) + Pr(Three Jacks) + Pr(Three Queens) + Pr(Three Kings) + Pr(Three Aces)
Now the probability of a three of a kind for any type of card is the same. Let’s just do Jacks:
Pr(Three of a kind for a Jack) = (4/20)*(4/20)*(4/20) = 1/125
So for all types of cards
P(three of a kind, any card) = Pr(Three 10s) + Pr(Three Jacks) + Pr(Three Queens) + Pr(Three Kings) + Pr(Three Aces)
= 5* Pr(Three of a kind for a Jack) = 5/125 = 1/25
The rest of the questions can be done with the same approach.
Richard | Jan 09, 2010
Since you replace the card each time, each draw is independent.
The probability of drawing any one card is (the number of cards available that fit the condition) divided by (the total number of cards) — in this case 20.
The probability of drawing a 10 [ p(10) ] is 4 out of 20 or 4/20 = 1/5 = 0.2
The probability of drawing an ace [ p(ace) ] is also 4 out fo 20 or 0.2
The probability of drawing both of these [ p(10 and ace) ] is 0.2 * 0.2 = 0.04
P(two hearts) = P(one heart) * P(one heart) = 5/20 * 5/20 = 1/4 * 1/4 = 1/16 = 0.0625
since there are 5 hearts available and 20 total cards.
P(three of any kind of card)
Imagine if you just wanted p(three aces), this would = 4/20 * 4/20 * 4/20 = 1/5 * 1/5 * 1/5
= 0.2*0.2*0.2 = .008 But there are 5 different types of cards available, so multiply .008 by 5 = .04
Understand?
paulcrsm | Jan 09, 2010
