1. ## Discrete Math question

Can someone help me with this?

2. ## Re: Discrete Math question

$\text{# of hands with at least 1 diamond} = \sum \limits_{k=1}^5 ~\text{# of hands with k diamonds}$

$\text{# of hands with k diamonds} = \dbinom{13}{k} \dbinom{39}{5-k}$

$\text{# of hands with at least 1 diamond} = \sum \limits_{k=1}^5~\dbinom{13}{k} \dbinom{39}{5-k} = 2023203$

There are a total of $\dbinom{52}{5}=2598960$ ways to deal a 5 card hand. Thus

$p = \dfrac{2023203}{2598960} = \dfrac{7411}{9520}$

Another, simpler way is to note that

$\text{# of hands with no diamonds} = \dbinom{39}{5} = 575757$

$\text{# of hands with at least 1 diamond} = \text{total # of hands}-\text{# of hands with no diamonds}=2598960-575757=2023203$

and we observe this is the same number computed by the series above.

3. ## Re: Discrete Math question

That seems to me the "hard way"! Instead I would calculate the probability of a hand with NO diamonds then subtract from 1.

Initially there are 52 cards, 3(13)= 36 of the "non-diamonds". The probability the first card drawn is not a diamond is 36/52= 3/4. Given that, there are 51 cards left, 35 of them "non-diamonds" so the probability the second card is not a diamond is 35/51= 11/17. Similarly the probability the third card is not a diamond is 34/50= 17/25, the probability the fourth card is not a diamond is 33/49, and the probability the fifth card is not a diamond is 32/48= 2/3.

The probability none of the cards in the hand is a diamond is (3/4)(11/17)(17/25)(33/49)(2/3) and the probability there is at least one diamond in the hand is 1 minus that.

4. ## Re: Discrete Math question

Originally Posted by HallsofIvy
That seems to me the "hard way"! Instead I would calculate the probability of a hand with NO diamonds then subtract from 1.
I did do both ways.

6. ## Re: Discrete Math question

Originally Posted by romsek
$\text{# of hands with at least 1 diamond} = \sum \limits_{k=1}^5 ~\text{# of hands with k diamonds}$

$\text{# of hands with k diamonds} = \dbinom{13}{k} \dbinom{39}{5-k}$

$\text{# of hands with at least 1 diamond} = \sum \limits_{k=1}^5~\dbinom{13}{k} \dbinom{39}{5-k} = 2023203$

There are a total of $\dbinom{52}{5}=2598960$ ways to deal a 5 card hand. Thus

$p = \dfrac{2023203}{2598960} = \dfrac{7411}{9520}$

Another, simpler way is to note that

$\text{# of hands with no diamonds} = \dbinom{39}{5} = 575757$

$\text{# of hands with at least 1 diamond} = \text{total # of hands}-\text{# of hands with no diamonds}=2598960-575757=2023203$

and we observe this is the same number computed by the series above.
Would this way be wrong?

Pr[drawing at least 1 diamond] = 1 - Pr[drawing 0 diamonds]
Number of cards other than diamonds = 52-13 = 39

Pr[drawing 0 diamonds] = C(39,5)/C(52,5)
Pr[drawing at least 1 diamond] = 1 - C(39,5)/C(52,5)
= 1- (575757 / 2598960)

= 0.7784663865546219