# 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.Yüklə 25,59 Kb.Dostları ilə paylaş:

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