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I have just started S2 and i am little stuck on this question, which appears to be simple but i am not to sure on my answer. The situation is:

A bag contains 4 red and 6 blue marbles. A marble is chosen at random, its colour is noted and it is then replaced in the bag. This procedure happens 11 times. Thanks in advance.

Sorry for not posting the question. It is in 2 parts.

What is the expected number of red marbles?

and

what is the most likely number of blue marbles drawn?

A bag contains 4 red and 6 blue marbles. A marble is chosen at random, its colour is noted and it is then replaced in the bag. This procedure happens 11 times. Thanks in advance.

Sorry for not posting the question. It is in 2 parts.

What is the expected number of red marbles?

and

what is the most likely number of blue marbles drawn?

What is the question? You haven't posted it

Since the marbles are replaced, the probability of a particular coloured marble is constant. Write down the probability of each and then tackle the question. Depending on the question, you might notice what kind of distribution it follows.

Since the marbles are replaced, the probability of a particular coloured marble is constant. Write down the probability of each and then tackle the question. Depending on the question, you might notice what kind of distribution it follows.

Okay I will set you off, as I have been told I give away answers too easy...

E(x) = np...

I suppose you can change the distirbution then work out each of the parts...

so you will use E(x) twice but with different np's...

PS if your still stuck or dont understand I will work out the answer for ya, do you have the answers btw so you can check yourself?

E(x) = np...

I suppose you can change the distirbution then work out each of the parts...

so you will use E(x) twice but with different np's...

PS if your still stuck or dont understand I will work out the answer for ya, do you have the answers btw so you can check yourself?

For the first question i got 4. But for the second because they said likely i put 7. As if they are 11 trials they has to be either 5 red drawn and 6 blue or 5 red drawn and 7 blue. so i assumed as the probability to draw blue is higher than red, it what more likely for there to be 7 blue drawn and 4 red. Am i right in thinking this?

"A bag contains 4 red and 6 blue marbles. A marble is chosen at random, its colour is noted and it is then replaced in the bag. This procedure happens 11 times."

"What is the expected number of red marbles?"

X~B(11, 0.4)

E(x) = np

E(x) = 11 x 0.4 = 4.4

"What is the most likely number of blue marbles drawn?"

Y~B(11, 0.6)

E(x) = np

E(x) = 11 x 0.6 = 6.6

Usually E(x) (Expected values of X) are left with decimals, and not rounded to the nearest marble.

"What is the expected number of red marbles?"

X~B(11, 0.4)

E(x) = np

E(x) = 11 x 0.4 = 4.4

"What is the most likely number of blue marbles drawn?"

Y~B(11, 0.6)

E(x) = np

E(x) = 11 x 0.6 = 6.6

Usually E(x) (Expected values of X) are left with decimals, and not rounded to the nearest marble.

Koudoo

"A bag contains 4 red and 6 blue marbles. A marble is chosen at random, its colour is noted and it is then replaced in the bag. This procedure happens 11 times."

"What is the expected number of red marbles?"

X~B(11, 0.4)

E(x) = np

E(x) = 11 x 0.4 = 4.4

"What is the most likely number of blue marbles drawn?"

Y~B(11, 0.6)

E(x) = np

E(x) = 11 x 0.6 = 6.6

Usually E(x) (Expected values of X) are left with decimals, and not rounded to the nearest marble.

"What is the expected number of red marbles?"

X~B(11, 0.4)

E(x) = np

E(x) = 11 x 0.4 = 4.4

"What is the most likely number of blue marbles drawn?"

Y~B(11, 0.6)

E(x) = np

E(x) = 11 x 0.6 = 6.6

Usually E(x) (Expected values of X) are left with decimals, and not rounded to the nearest marble.

Thanks. That is what i initially did, but then i thought in real terms that you couldnt remove 6.6 marbles or 4.4. I was forgetting that this is statistics and it is theoretical.

Koudoo

Do you have the answers?

Your thought process was right, but I don't think it would achieve you the full marks, since there is a direct method to solving the question.

Your thought process was right, but I don't think it would achieve you the full marks, since there is a direct method to solving the question.

Well it is multiple choice, but i dont have the answer sheet. The options for the first question are:

4, 4.4, 5 and 6.6

for th 2nd they are:

7, 6, 6.6 and 4.

Koudoo

Do they want the statistical answer or the nearest marble answer?

As thats the hardest thing about these questions, as I am never sure when you are expected to relate it back to the question and give a real answer, or leave it as a statistical answer ...

As thats the hardest thing about these questions, as I am never sure when you are expected to relate it back to the question and give a real answer, or leave it as a statistical answer ...

Yh that was the problem i encountered. But if we go back to the theory, the expected value is not always a possible outcome. So following the rules of statistics i think that your working out is correct. Because they havent said whether i should relate it to the real world.

At all the answer schemes I seen a long while back when I done it, they usually want the decimal/orginal answer, but you can get the odd question.

But I think the big trigger word is where they said "Expected value", and that also some of the choice options had decimals in...

I take it this isnt a exam question?

But I think the big trigger word is where they said "Expected value", and that also some of the choice options had decimals in...

I take it this isnt a exam question?

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