Question

Please answer I’ll rate brainlyest

Answer 1

Answer:

D

Step-by-step explanation:

We can choose 3 cards from 52 cards in $C_3^{52}$ different ways.

1. There are 4 aces and 4 tens. To select 2 aces form 4 we have $C^4_2$ different ways and to select 1 ten from 4 we have $C^4_1$ different ways. Thus, the probability

$Pr(\text{2 aces and 1 ten})=\dfrac{C^4_2\cdot C^4_1}{C^{52}_3}.$

This option is false.

2. There are 4 fives. To select 2 fives form 4 we have $C^4_2$ different ways. But we have to choose 3 cards, so the remaining 3rd card must be chosen from the rest 48 cards in $C^{48}_1$ different ways. Thus, the probability

$Pr(\text{2 fives})=\dfrac{C^4_2\cdot C^{48}_1}{C^{52}_3}.$

This option is false.

3. There are 13 hearts and 13 spades. To select 1 heart form 13 we have $C^{13}_1$ different ways and to select 2 spades from 13 we have $C^{13}_2$ different ways. Thus, the probability

$Pr(\text{1 heart and 2 spades})=\dfrac{C^{13}_1\cdot C^{13}_2}{C^{52}_3}.$

This option is false.

4. There are 4 kings. To select no kings we have $C^{48}_3$ different ways . Thus, the probability

$Pr(\text{no kings})=\dfrac{C^{48}_3}{C^{52}_3}.$

This option is true.