How many ways are there of choosing a hand of 5 cards with at least 3 red cards from a deck of 52 cards?

Question

Standard deck of cards with 26 red cards and 26 black ones.

Is it ${26 \choose 3} \times {26 \choose 2} + {26 \choose 4} \times {26 \choose 1} + {26 \choose 5} \times {26 \choose 0}$?

Answer 1

Yes. Out of 52 cards, there are 26 red and 26 black cards. So the possibilities of choosing a hand of 5 cards with at least 3 red cards are summation of:

3 Red 2 Black : $^{26}C^{}_{3}\cdot^{26}C^{}_{2}$

4 Red 1 Black : $^{26}C^{}_{4}\cdot^{26}C^{}_{1}$

5 Red 0 Black : $^{26}C^{}_{5}\cdot^{26}C^{}_{0}$

$Ans=^{26}C^{}_{3}\cdot^{26}C^{}_{2}+^{26}C^{}_{4}\cdot^{26}C^{}_{1}+^{26}C^{}_{5}\cdot^{26}C^{}_{0}$

$Ans=2600\cdot325+14950\cdot26+65780\cdot1=1299480$

Answer 2

Here's another approach. Each 5-card hand will either have at least three red cards or at least 3 black cards, but not both. Since there is no preference between black and red cards (since they are equal in number in the deck), in exactly half of all cases you will get at least 3 reds.

So the answer is $\frac{1}{2}\cdot{{52}\choose{5}}$

This gives the same answer as Naman Jain's answer.