# Math Help - Probability

1. ## Probability

A card deck has 52 cards, a person will randomly choose four card with replacement. Find the probability that the person will get at least 3 hearts?

Please help as I don't understand what is with replacement and will order matter when we choose exactly 3 hearts

2. Pick a card. You get 1 of 52 possibilities.

If you put the card back, the next draw remains 1 of 52 possibilities.

If you do NOT put the card back, the next draw is 1 of 51 possibilities.

Put it back? With Replacement.

Don't Put it back? Without Replacement.

Each draw is a heart or it isn't. WITH replacement, the probabilties do not change with each draw.

13/52 heart 39/52 not a heart

If one is to get ALL FOUR hearts, that shoudl be (13/52)^4. There is only one sequence that will produce this result.

If one is get THREE hearts, there are four different sequences that will work. OHHH, HOHH, HHOH, and HHHO. Each of these has probability (39/52)*(13/52)^3. Multiplication is commutative, so the probablities don't change. The order DOES matter a little - it makes four ways to do it, rather than just one.

What do you get?

3. Originally Posted by TKHunny
Pick a card. You get 1 of 52 possibilities.

If you put the card back, the next draw remains 1 of 52 possibilities.

If you do NOT put the card back, the next draw is 1 of 51 possibilities.

Put it back? With Replacement.

Don't Put it back? Without Replacement.

Each draw is a heart or it isn't. WITH replacement, the probabilties do not change with each draw.

13/52 heart 39/52 not a heart

If one is to get ALL FOUR hearts, that shoudl be (13/52)^4. There is only one sequence that will produce this result.

If one is get THREE hearts, there are four different sequences that will work. OHHH, HOHH, HHOH, and HHHO. Each of these has probability (39/52)*(13/52)^3. Multiplication is commutative, so the probablities don't change. The order DOES matter a little - it makes four ways to do it, rather than just one.

What do you get?

Thanks but I just do not seem to understand why not

((39/52) x (13/52)^3) x4

Why not times 4 as the the order can be different

4. ## Probability

'At least' three hearts means three or more hearts. So the HHHH combination must also be considered.

I hope that helps.

ILoveMaths07.

5. It's actually like this, someone.

$P$ (picking at least three hearts) $= P (HHHH) + P (HHHN) + P (NHHH) + P (HNHH) + P (HHNH)$

$= \left(\frac {13}{52}\right)^4 + 4 * \left(\frac {13}{52}\right)^3 * \left(\frac{39}{52}\right)$

$= \frac {1}{256} + 4 * \frac {3}{256}$

$= \frac {1}{256} + \frac {12}{256}$

$= \frac {13}{256}.$

I hope that helps.

ILoveMaths07.

6. Originally Posted by ILoveMaths07
'At least' three hearts means three or more hearts. So the HHHH combination must also be considered.
Yup. I missed that.

7. Thanks everyone for helping still one question

As seen from this question order does matter so is there any rules for when to consider order??

8. Umm... lol I know what you mean. I've been through the same mathematical dilemma. At times, the order matters... At times, it doesn't. All I can say is PRACTISE A LOT. Then you'll get a feel for it, and you'll know when order matters, and when it does not.

In this question, it DOES matter because the hearts can be selected in any of those ways. Drawing a heart on your first three goes and not drawing it on your last is NOT the same as drawing any other suit on your first go and drawing a heart on your next three. Yes, the probabilities of these happening are the same, but the events are different. Events that are different must be considered. Events that are the same must not be considered.
For e.g. You might argue that... HHHH also has many other possibilities that depend on the order of cards drawn. For instance,
1, 2, 3, 4
4, 1, 2, 3
J, Q, K, A
Q, J, A, K
2, 7, 9, 8
2, 9, 8, 7, and tonnes of others.
But they are basically THE SAME. Because you're looking at whether the card is a heart or not. Drawing 1, 2, 3, 4 is the same as drawing 2, 1, 4, 3 or even the same as drawing J, Q, K, A AS LONG AS THEY'RE ALL HEARTS! So you consider HHHH only, and not several other possibilities.

I hope that gives you an insight into when to consider the order as important, and when not to.

I hope that helps.

ILoveMaths07.

9. had to bring it again

but why when we choose at least 3 hearts, for the 3 hearts part do we really have to apply order, thereby multiple by 4.

10. ## Prob

someone, in the case of three hearts, you have to consider order because the fourth non-heart card can be drawn on your first, second, third, or fourth go. There are four different possibilities, and hence, four events.
In the case of all four hearts, you do not have to consider order because we do not care what heart it is. Whether it's an Ace or a 2 doesn't make a difference... as long as it is a heart. The question does not ask about picture cards, court cards etc. The question asks about hearts only. So, if all four are hearts, you do not have to worry about which cards you drew!

I hope that helps.

me07.

11. Originally Posted by ILoveMaths07
someone, in the case of three hearts, you have to consider order because the fourth non-heart card can be drawn on your first, second, third, or fourth go. There are four different possibilities, and hence, four events.
In the case of all four hearts, you do not have to consider order because we do not care what heart it is. Whether it's an Ace or a 2 doesn't make a difference... as long as it is a heart. The question does not ask about picture cards, court cards etc. The question asks about hearts only. So, if all four are hearts, you do not have to worry about which cards you drew!

I hope that helps.

me07.
I am trusting you and am going with order even through my friends and many other people dont agree

I will definately post the correct answer for this question when I recieve them as it surely is for me one of the most frustrating questions

12. A good way to get around this (Why didn't I think of this before? ) is to draw a tree diagram.

If I'm not wrong, you will get 16 combinations in the end... 5 of which will contain 3 hearts or more - HHHH, NHHH, HNHH, HHNH and HHHN.

I hope that helps.

ILoveMaths07.

13. Originally Posted by ILoveMaths07
A good way to get around this (Why didn't I think of this before? ) is to draw a tree diagram.

If I'm not wrong, you will get 16 combinations in the end... 5 of which will contain 3 hearts or more - HHHH, NHHH, HNHH, HHNH and HHHN.