Question

] Consider a randomly shuffled deck of ten cards labeled {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three to be drawn. Which one should be worth more points: A getting a pair of matching cards within the three, or B getting a set of three cards that can be arranged such that the sum of two of them is the value of the third (e.g. 1 2

Answer 1

Answer:

B should be worth more points.

Step-by-step explanation:

Given -  Consider a randomly shuffled deck of ten cards labeled

              {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three

               to be drawn.

To find - Which one should be worth more points:

              A getting a pair of matching cards within the three,

              or B getting a set of three cards that can be arranged such

              that the sum of two of them is the value of the third.

Proof -

Given that, three cards has to be drawn.

So, total number of ways the card be drawn = ¹⁰C₃

                                                                      = $\frac{10!}{3! (10-3)!}$

                                                                      = $\frac{10!}{3! 7!}$

                                                                      = $\frac{10.9.8.7!}{3! 7!}$

                                                                      = $\frac{10.9.8}{3.2.1}$

                                                                      = 120

⇒¹⁰C₃ = 120

Now,

Given that, A getting a pair of matching cards within the three

Sample space = { (1,1,2), (1,1,3).(1,1,4),(1,1,5);

                            (2,2,1),(2,2,3),(2,2,4),(2,2,5);

                             (3,3,1),(3,3,2),(3,3,4),(3,3,5);

                              (4,4,1),(4,4,2),(4,4,3),(4,4,5);

                              (5,5,1),(5,5,2),(5,5,3),(5,5,4) }

⇒n(A) = 20

Also,

B getting a set of three cards that can be arranged such that the sum of two of them is the value of the third

Sample space = { (1,1,2),(1,2,3),(1,3,4),(1,4,5),(2,2,4),(2,3,5) }

⇒n(B) = 6

Now,

P(A) = $\frac{20}{120} = \frac{1}{6}$ = 0.167

P(B) = $\frac{6}{120} = \frac{1}{20}$ = 0.05

We can see that, P(B) < P(A)

⇒B should be worth more points.