[MOVED] Probability of drawing same number in card game

[Monkeyseat]

10) Jodie and Karen play a game with a pack of 40 numbered cards.
There are four cards with each numbers 0 to 9.

Jodie picks a card and keeps it. Karen then picks a card.

a) What is the probability they both pick a number 5?

4/40 * 3/39 = 12/1560 = 3/390 = 1/130 (is that correct?)

b) After each game they replace the cards.

They play the game 50 times. They each pick a card with the same number in 12 of the games. Is this more than you would expect or less? Explain your answer clearly, showing your calculations.

Game 1:

1/130 * 50 = 5/13

Game 2:

12/50

Get same denominators:

Game 1:

250/650 = 125/325

Game 2:

156/650 = 78/325

Less likely.

I've probably done that wrong, so can someone give me some help please?

Thank you very much.

 

[soroban]

Hello, Monkeyseat!

Ya done good!

10) Jodie and Karen play a game with a pack of 40 numbered cards.
There are four cards with each numbers 0 to 9.

Jodie picks a card and keeps it. .Karen then picks a card.

a) What is the probability they both pick a number 5?

\(\displaystyle \frac{4}{40}\cdot\frac{3}{39}\:=\:\frac{1}{10}\cdot\frac{1}{13}\:=\: \frac{1}{130}\;\) (is that correct?)

Yes!

b) After each game they replace the cards.
They play the game 50 times.
They each pick a card with the same number in 12 of the games.

Is this more than you would expect or less?
Explain your answer clearly, showing your calculations.

You are correct!

I would format the answer like this:

. . \(\displaystyle \begin{array}{cccccc}\text{Expected: }& \frac{1}{130}(50) & \:=\: & \frac{5}{13} & \:=\:& \frac{125}{325} \\ \\ \\

\text{Actual: } &\frac{12}{50} & = & \frac{6}{25} & = & \frac{78}{325}\end{array}\)


Therefore: \(\displaystyle \:\text{Actual }\:<\:\text{ Expected}\)

 

[pka]

Note that for part (b) the drawing is a match not necessarily a five.
The first card can be anyone of 40.
Then the probability of the second card matching is \(\displaystyle \frac{3}{{39}} = \frac{1}{{13}}\).

By replacing the cards after each round, we have a binominal experiment.
The expected number is \(\displaystyle n \cdot p\) where \(\displaystyle n\) is the numbers of trials and \(\displaystyle p\) is the probability of the event.

In fifty draws what is the expected number of matches?

 

[Monkeyseat]

Thanks for the replys guys.

Ha, I didn't think it was right. So is game 1 the "Prediction" of the expected results, and B the actual results?

I'm not quite sure what you're saying pka, that they don't have to be a 5 in game 2? I think you have to assume it really from A (as it is asking is this more or less likely), although I'm not entirely sure. What is the answer to b then? :?

 

[pka]

You have to assume it really from A (as it is asking is this more or less likely), What is the answer to b then? :?

There is absolutely no reason to assume that part b follows from part a.
Read it very carefully: After each game they replace the cards.
They play the game 50 times. They each pick a card with the same number in 12 of the games. Is this more than you would expect or less?
It says the same number; it does not say the card having a 5 on it!

Thus on each game the probability is \(\displaystyle {\frac{1}{13}}\) for both getting the same number on the cards. In fifty games, the expected number of times that it should happen is \(\displaystyle (50){\frac{1}{13}}\)= \(\displaystyle {\frac{50}{13}} \propto 4\) . That is far less than the observed 12.

 

[Monkeyseat]

You have to assume it really from A (as it is asking is this more or less likely), What is the answer to b then? :?

There is absolutely no reason to assume that part b follows from part a.
Read it very carefully: After each game they replace the cards.
They play the game 50 times. They each pick a card with the same number in 12 of the games. Is this more than you would expect or less?
It says the same number; it does not say the card having a 5 on it!

Thus on each game the probability is \(\displaystyle {\frac{1}{13}}\) for both getting the same number on the cards. In fifty games, the expected number of times that it should happen is \(\displaystyle (50){\frac{1}{13}}\)= \(\displaystyle {\frac{50}{13}} \propto 4\) . That is far less than the observed 12.

I see, but why is it 1/13 exactly? I thought it was 1/130 after 1 game?

After 50 games it's 5/13 isn't it, which is more likey than 12/50...?

Thanks.

And where did you get the directly proportional to 4 from?

Edit:

I get it now:

1 (because you can choose any card the first time) times 3/39 is 1/13.

But why is it proportional to 4? And isn't 50/13 a lot more than 12/50? Put that in bold because that's the main bit I don't get.

 

[pka]

You seem to be saying that in part (b) they both drew a five twelve times in fifty plays. If that is what the question means then the expected value of that happing in fifty independent trials is \(\displaystyle (50){\frac{1}{130}} \propto 0.4\) or less than 1 time. That is far far less than the observed 12.

 

[Monkeyseat]

You seem to be saying that in part (b) they both drew a five twelve times in fifty plays. If that is what the question means then the expected value of that happing in fifty independent trials is \(\displaystyle (50){\frac{1}{130}} \propto 0.4\) or less than 1 time. That is far far less than the observed 12.

No I understand what you mean now.

But why is it proportional to 4?

And isn't 50/13 a lot more than 12/50? Please just say when it is less than observed (it's not meant to be 1/13 x 1/13 x 1/13 etc. to 50 is it?)...

50/13 = 3.85

12/50 = 0.24

So isn't the probability of their actual outcome a lot less than it is meant to be? Is 3.85 an actual possiblilty (does it not have to be below 1?)?

 

[pka]

Let me give you a very basic lesson in expected value.
Think if a die. Roll it. The probability of getting a five is \(\displaystyle {\frac{1}{6}}\).
If you roll the die fifty times the expected number of times that a 5 comes up is \(\displaystyle (50){\frac{1}{6}} = {\frac{50}{6}} \propto 8\). That is, fifty rolls of a die yields on average 8 fives. (For that matter, the number 1 ought to appear about 8 times.)

Take another example. There is a box with 10 marbles, 9 white and one red.
We draw one marble, note its color and put it back. If we do that 30 times, how many of the picks do we expect to have been red?
Answer: 3, that is \(\displaystyle (30){\frac{1}{10}} = 3\).

If the probability of both girls drawing a 5 is \(\displaystyle {\frac{1}{130}}\) then what is the expected numbers of times that they will both draw a 5 is fifty rounds?

 

[Monkeyseat]

Let me give you a very basic lesson in expected value.
Think if a die. Roll it. The probability of getting a five is \(\displaystyle {\frac{1}{6}}\).
If you roll the die fifty times the expected number of times that a 5 comes up is \(\displaystyle (50){\frac{1}{6}} = {\frac{50}{6}} \propto 8\). That is, fifty rolls of a die yields on average 8 fives. (For that matter, the number 1 ought to appear about 8 times.)

Take another example. There is a box with 10 marbles, 9 white and one red.
We draw one marble, note its color and put it back. If we do that 30 times, how many of the picks do we expect to have been red?
Answer: 3, that is \(\displaystyle (30){\frac{1}{10}} = 3\).

If the probability of both girls drawing a 5 is \(\displaystyle {\frac{1}{130}}\) then what is the expected numbers of times that they will both draw a 5 is fifty rounds?

Where did the 4 come from in 50/13 (why exactly is it proportional, they are linked/mean the same?)? Why does it yield 4 times? The probability of winning 1/13 is 0.07692... times 50 is 3.84... Is it because 3.84... rounds up to 4? You do that because they are independant, otherwise you would multiply them together?

If you were to write it as a possibility what would it be i.e. 5/12 is 0.24 being 24%? Does that mean percent or the amount it should yield? I think it's just 0.07692 because:

0.24/0.0796 is 3.12. 3.84 * 3.12 is 12.

I've got an exam tomorrow and I've got to go out in a minute so I'm trying to get this ASAP.

Apologies for the amount of questions, but it would be great if you could answer them and clear it up for me. Sorry for taking a while lol to get it, I get it a bit now.
Thanks.