Answer:
D
Step-by-step explanation:
We can choose 3 cards from 52 cards in [tex]C_3^{52}[/tex] different ways.
1. There are 4 aces and 4 tens. To select 2 aces form 4 we have [tex]C^4_2[/tex] different ways and to select 1 ten from 4 we have [tex]C^4_1[/tex] different ways. Thus, the probability
[tex]Pr(\text{2 aces and 1 ten})=\dfrac{C^4_2\cdot C^4_1}{C^{52}_3}.[/tex]
This option is false.
2. There are 4 fives. To select 2 fives form 4 we have [tex]C^4_2[/tex] different ways. But we have to choose 3 cards, so the remaining 3rd card must be chosen from the rest 48 cards in [tex]C^{48}_1[/tex] different ways. Thus, the probability
[tex]Pr(\text{2 fives})=\dfrac{C^4_2\cdot C^{48}_1}{C^{52}_3}.[/tex]
This option is false.
3. There are 13 hearts and 13 spades. To select 1 heart form 13 we have [tex]C^{13}_1[/tex] different ways and to select 2 spades from 13 we have [tex]C^{13}_2[/tex] different ways. Thus, the probability
[tex]Pr(\text{1 heart and 2 spades})=\dfrac{C^{13}_1\cdot C^{13}_2}{C^{52}_3}.[/tex]
This option is false.
4. There are 4 kings. To select no kings we have [tex]C^{48}_3[/tex] different ways . Thus, the probability
[tex]Pr(\text{no kings})=\dfrac{C^{48}_3}{C^{52}_3}.[/tex]
This option is true.
