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Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 5"

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== Solution ==
 
== Solution ==
 +
<math>\frac{2}{15}</math>
  
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There are five ways to achieve a sum divisible by 7; 115 (2 ways), 133 ( 2 ways), 124 (8 ways), 223 (2 ways), 455 (2 ways). Hence, there are 16 favorable ways out of <math>120=\binom{10}{3}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:56, 25 July 2023

Problem

Ten different playing cards have the numbers $1, 1, 2, 2, 3, 3, 4, 4, 5, 5$ written on them as shown. Three cards are selected at random without replacement. What is the probability that the sum of the numbers on the three cards is divisible by $7$?

[asy] size(200,100); path Card=(arc((.2,.2),.2,180,270)--(1.8,0)--arc((1.8,.2),.2,270,360)--(2,2.8)--arc((1.8,2.8),.2,0,90)--(.2,3)--arc((.2,2.8),.2,90,180)--(0,.2)); draw(Card,black); draw(shift(3,0)*Card,black); draw(shift(6,0)*Card,black); draw(shift(9,0)*Card,black); draw(shift(12,0)*Card,black); draw(shift(0,-4)*Card,black); draw(shift(3,-4)*Card,black); draw(shift(6,-4)*Card,black); draw(shift(9,-4)*Card,black); draw(shift(12,-4)*Card,black); MP("1",(.25,2.1),N);MP("1",(1.75,-.1),N); MP("1",(.25,2.1-4),N);MP("1",(1.75,-4.1),N); MP("\spadesuit",(1,1),N);MP("\clubsuit",(1,1-4),N); MP("2",(3.25,2.1),N);MP("2",(4.75,-.1),N); MP("2",(3.25,2.1-4),N);MP("2",(4.75,-4.1),N); MP("\spadesuit",(4,1.75),N);MP("\clubsuit",(4,1.75-4),N); MP(180,"\spadesuit",(4,.25),N);MP(180,"\clubsuit",(4,.25-4),N); MP("3",(6.25,2.1),N);MP("3",(7.75,-.1),N); MP("3",(6.25,2.1-4),N);MP("3",(7.75,-4.1),N); MP("\spadesuit",(7,1.75),N);MP("\clubsuit",(7,1.85-4),N); MP("\spadesuit",(7,1),N);MP("\clubsuit",(7,1-4),N); MP(180,"\spadesuit",(7,.25),N);MP(180,"\clubsuit",(7,.25-4),N); MP("4",(9.25,2.1),N);MP("4",(10.75,-.1),N); MP("4",(9.25,2.1-4),N);MP("4",(10.75,-4.1),N); MP("\spadesuit",(10-.3,1.75),N);MP("\clubsuit",(10-.3,1.75-4),N); MP(180,"\spadesuit",(10-.3,.25),N);MP(180,"\clubsuit",(10-.3,.25-4),N); MP("\spadesuit",(10+.3,1.75),N);MP("\clubsuit",(10+.3,1.75-4),N); MP(180,"\spadesuit",(10+.3,.25),N);MP(180,"\clubsuit",(10+.3,.25-4),N); MP("5",(12.25,2.1),N);MP("5",(13.75,-.1),N); MP("5",(12.25,2.1-4),N);MP("5",(13.75,-4.1),N); MP("\spadesuit",(13-.3,1.75),N);MP("\clubsuit",(13-.3,1.75-4),N); MP("\spadesuit",(13,1),N);MP("\clubsuit",(13,1-4),N); MP(180,"\spadesuit",(13-.3,.25),N);MP(180,"\clubsuit",(13-.3,.25-4),N); MP("\spadesuit",(13+.3,1.75),N);MP("\clubsuit",(13+.3,1.75-4),N); MP(180,"\spadesuit",(13+.3,.25),N);MP(180,"\clubsuit",(13+.3,.25-4),N); [/asy]


Solution

$\frac{2}{15}$

There are five ways to achieve a sum divisible by 7; 115 (2 ways), 133 ( 2 ways), 124 (8 ways), 223 (2 ways), 455 (2 ways). Hence, there are 16 favorable ways out of $120=\binom{10}{3}$.

See Also

2007 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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