Respuesta :
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
= [tex]\frac{10!}{3! (10-3)!}[/tex]
= [tex]\frac{10!}{3! 7!}[/tex]
= [tex]\frac{10.9.8.7!}{3! 7!}[/tex]
= [tex]\frac{10.9.8}{3.2.1}[/tex]
= 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) = [tex]\frac{20}{120} = \frac{1}{6}[/tex] = 0.167
P(B) = [tex]\frac{6}{120} = \frac{1}{20}[/tex] = 0.05
We can see that, P(B) < P(A)
⇒B should be worth more points.
