cooloicons62

2022-07-13

Combinatorics select 3 cards out of 52, probability all 3 spades and not.

You get 3 cards out of 52, the order doesn't matter.

a) What is the probability all 3 cards are spades?

b) What is the probability none of the 3 cards are spades?

a) I think it is

$\frac{{\textstyle (}\genfrac{}{}{0ex}{}{13}{3}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}\approx 0.012941$

b) I know it is

$=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{39}{3}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}\approx 0.41353$

I know $P({A}^{c})=1-P(A)$

And if a is $A$ and b is ${A}^{c}$ the equation above is not correct..

You get 3 cards out of 52, the order doesn't matter.

a) What is the probability all 3 cards are spades?

b) What is the probability none of the 3 cards are spades?

a) I think it is

$\frac{{\textstyle (}\genfrac{}{}{0ex}{}{13}{3}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}\approx 0.012941$

b) I know it is

$=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{39}{3}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}\approx 0.41353$

I know $P({A}^{c})=1-P(A)$

And if a is $A$ and b is ${A}^{c}$ the equation above is not correct..

Karla Hull

Beginner2022-07-14Added 20 answers

Let P(n) be the probability that n cards are drawn. P(3) is the answer for a), and you got that right. The answer for b) is P(0).

$P(0)+P(1)+P(2)+P(3)=1$, so $P(0)+P(3)<1$

$P(n)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{13}{n}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{39}{3-n}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}$

$P(0)+P(1)+P(2)+P(3)=1$, so $P(0)+P(3)<1$

$P(n)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{13}{n}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{39}{3-n}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{3}{\textstyle )}}$

Ximena Skinner

Beginner2022-07-15Added 7 answers

The issue here is that the complement of the event "all of the 3 cards are spades" is "at least one of the 3 cards is not a spade" not "none of the 3 cards are spades".

So, $P(\u2018\u2018\text{all of the 3 cards are spades}")+P(\u2018\u2018\text{none of the 3 cards are spades}")<1$

So, $P(\u2018\u2018\text{all of the 3 cards are spades}")+P(\u2018\u2018\text{none of the 3 cards are spades}")<1$

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