Pregunta Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and y the number of jacks



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Pregunta

Pregunta

Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and Y the number of jacks. Find (a) the joint probability distribution of X and Y; (b) P[(X, Y) in A, where A is the region given by

(x,y)∣x+y≥2.{(x, y) | x + y \geq 2 }.(x,y)∣x+y≥2.

Explicación

Verificado



Paso 1

1 de 4


(a)\textbf{(a)}(a)

Since there are 4 kings and 4 jacks overall, and 3 cards are being drawn, we know that the random variables

XXX

and


YYY

can assume values 0,1,2 and 3.

We can immediately conclude:

P(X=3,Y=1)=P(X=3,Y=2)=P(X=3,Y=3)=0\boxed{ P(X=3, Y=1) = P(X=3, Y=2) = P(X=3, Y=3) =0 }P(X=3,Y=1)=P(X=3,Y=2)=P(X=3,Y=3)=0​

P(X=2,Y=2)=P(X=2,Y=3)=P(X=1,Y=3)=0\boxed{ P(X=2, Y=2) = P(X=2, Y=3) = P(X=1, Y=3) = 0 }P(X=2,Y=2)=P(X=2,Y=3)=P(X=1,Y=3)=0​

Those probabilities are 0 because those events suggest that a

total\textbf{total}total

of


more than 3‾\text{\underline{more than 3}}more than 3​

cards was drawn which is incorrect.

All other combinations of values, which the random variables

XXX


and

YYY


are able to assume, are possible.

We notice that, if the draw contains

xxx

kings and



yyy

jacks, it also contains

3−x−y3-x-y3−x−y

queens. Therefore, the number of ways in which

xxx

kings and



yyy

jacks can be chosen into the sample is

(4x)(4y)(43−x−y)\binom{4}{x} \binom{4}{y} \binom{4}{3-x-y}(x4​)(y4​)(3−x−y4​)

for


(x,y) such that x+y≤3(x,y) \text{ such that } x+y \leq 3(x,y) such that x+y≤3

.

Having in mind that the



total\textbf{total}total

number of ways in which a draw of 3 cards among 12 can be selected is

(123)\binom{12}{3}(312​)

, we find the formula:

f(x,y)=P(X=x,Y=y)=(4x)(4y)(43−x−y)(123)\boxed{ f(x,y) = P(X=x,Y=y) = \frac{\binom{4}{x} \binom{4}{y} \binom{4}{3-x-y}}{\binom{12}{3}} }f(x,y)=P(X=x,Y=y)=(312​)(x4​)(y4​)(3−x−y4​)​​

for


(x,y) such that x+y≤3(x,y) \text{ such that } x+y \leq 3(x,y) such that x+y≤3

.

Paso 2

2 de 4



Paso 3

3 de 4


(b)\textbf{(b)}(b)

Taking into account the joint probability distribution table, we get:

P((X,Y)∈A)=P(X+Y≥2)=1−P(X+Y<2)=1−P(X+Y≤1)=1−(P(X=0,Y=0)+P(X=1,Y=0)+P(X=0,Y=1))=1−(155+655+655)=1−1355=4255\begin{align*} P\bigr( (X,Y) \in A \bigr) &= P(X+Y \geq 2) \\ &= 1 - P(X+Y <2) \\ &= 1 - P(X+Y \leq 1) \\ &= 1 - \bigr( P(X=0, Y=0) + P(X=1, Y=0) + P(X=0, Y=1) \bigr) \\ &= 1 - \biggr( \frac{1}{55} + \frac{6}{55} + \frac{6}{55} \biggr) \\ &= 1 - \frac{13}{55} \\ &= \boxed{ \frac{42}{55} } \end{align*}P((X,Y)∈A)​=P(X+Y≥2)=1−P(X+Y<2)=1−P(X+Y≤1)=1−(P(X=0,Y=0)+P(X=1,Y=0)+P(X=0,Y=1))=1−(551​+556​+556​)=1−5513​=5542​​​

Resultado

4 de 4


See explanation.
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