probability problem

[allegansveritatem]

I came across this example in my precalculus book:


I went through this carefully and then decided to see if I could apply it to a similar problem, namely instead of looking for 5 hearts, I asked what the probability would be for being dealt 4 hearts instead of 5. I figured that the probability would be greater but when, using this procedure, I tried to find this probability I got this:


So, what is going on here? How would I have to adjust the denominator (I am guessing) to get a probability that makes sense here?

 

[Dr.Peterson]

I came across this example in my precalculus book:
View attachment 25179

I went through this carefully and then decided to see if I could apply it to a similar problem, namely instead of looking for 5 hearts, I asked what the probability would be for being dealt 4 hearts instead of 5. I figured that the probability would be greater but when, using this procedure, I tried to find this probability I got this:
View attachment 25182

So, what is going on here? How would I have to adjust the denominator (I am guessing) to get a probability that makes sense here?

If you draw only 4 cards, doesn't that change the sample space (denominator)?

 

[Steven G]

I think that you still want to draw 5 cards but only get 4 hearts. You forgot to calculate the number of ways to draw that last card. Do you do hypergeometric problem yet?

 

[allegansveritatem]

If you draw only 4 cards, doesn't that change the sample space (denominator)?

well, not really, because I am still drawing a 5 card hand. And, I believe I tried this problem using C(52,4) and come up with a probability that was still not smaller than what it is for drawing 5 hearts. I will ponder this today and see what I can come up with.

 

[allegansveritatem]

I think that you still want to draw 5 cards but only get 4 hearts. You forgot to calculate the number of ways to draw that last card. Do you do hypergeometric problem yet?

I do not even know what a hypergeometric problem is but I will look it up and get back.

 

[Dr.Peterson]

well, not really, because I am still drawing a 5 card hand. And, I believe I tried this problem using C(52,4) and come up with a probability that was still not smaller than what it is for drawing 5 hearts. I will ponder this today and see what I can come up with.

Your numerator implies you are only drawing 4 cards, so I got in my mind that was what you were solving.

Since you are drawing 5 cards, the numerator has to account for that. So, how many ways are there to draw 4 hearts AND a non-heart? Give it a try.

 

[allegansveritatem]

Your numerator implies you are only drawing 4 cards, so I got in my mind that was what you were solving.

Since you are drawing 5 cards, the numerator has to account for that. So, how many ways are there to draw 4 hearts AND a non-heart? Give it a try.

I will try it and post results later. Thanks

 

[allegansveritatem]

Thanks to the epiphanous (?) advice of Dr Peterson I was able, with the help of my Casio 991 emulator, to come to terms with this exercise thus: