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A card is chosen from a well-shuffled standard deck of 52 cards. What is the probability (percent chance) that the card will be 4 or an Even card? Find P(4 or Even)

 A.38.46 %

B.30.77 %

C.2.37 %

D.46.15 %

 

Update****  Those are the choices for the answers.

 Dec 12, 2018
edited by so1963  Dec 12, 2018

Best Answer 

 #2
avatar+1368 
+3

Solution:

 

\(\text {A quick method for solving is to note that all fours (4) are even and }\\ \text { cannot be independent of even, so the probability is } \dfrac {6}{13} = .4615 \\ \text { }\\ \text {Formally: }\\ \large \mathbb \rho \small (4 \text { or even}) = \large \mathbb \rho \small (4) + \large \mathbb \rho \small (\text {even}) - \large \mathbb \rho \small (4 \text { and even})\\ = \dfrac{4}{52} + \dfrac {24}{52} - \left( \dfrac {4}{52} * 1 \right) = \mathbf {46.15} \%\\ \)

 

 

GA

 Dec 12, 2018
 #1
avatar+98102 
+1

**********

 

 

cool cool cool

 Dec 12, 2018
edited by CPhill  Dec 12, 2018
 #2
avatar+1368 
+3
Best Answer

Solution:

 

\(\text {A quick method for solving is to note that all fours (4) are even and }\\ \text { cannot be independent of even, so the probability is } \dfrac {6}{13} = .4615 \\ \text { }\\ \text {Formally: }\\ \large \mathbb \rho \small (4 \text { or even}) = \large \mathbb \rho \small (4) + \large \mathbb \rho \small (\text {even}) - \large \mathbb \rho \small (4 \text { and even})\\ = \dfrac{4}{52} + \dfrac {24}{52} - \left( \dfrac {4}{52} * 1 \right) = \mathbf {46.15} \%\\ \)

 

 

GA

GingerAle Dec 12, 2018
 #3
avatar+98102 
0

(Facepalm)......that was too easy, GA......

 

I should have consulted one of your learned  chimps.....even THEY could have done better than I did!!!!

 

 

 

cool cool cool

CPhill  Dec 12, 2018

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