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Veseljchak [2.6K]
3 years ago
6

Please answer I’ll rate brainlyest

Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

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.

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