Combinations question (1 Viewer)

[enigma_1]

I don't have solutions to this, that's why

Question: There are 3 red and 2 white balls in a bag. A boy selects 2 balls, one after the other.

i) Find the probability that the two balls are white
ii) Find probability that at least one ball is red
iii) After selecting 2 balls, one ball slipped out of his hand which is red. What is the probability that the other ball in his hand is also red?

Thanks heaps

 

[braintic]

This is an HSC question, and the last part has caused much disagreement.

The correct answer is 1/3. The answer that was marked correct at the time was 1/5.
MANSW took 10 years to change their official answer to 1/3.
Despite the change, many teachers refuse to teach it correctly.

The people who get 1/5 are answering the wrong question ... that answer IS correct for a slightly modified version of the question.
As it stands, the question is VERY easy, but some people choose to overthink the question.

But in the end, the question should not have been asked ... it is conditional probability, which is not in the course.

 

[enigma_1]

This is an HSC question, and the last part has caused much disagreement.

The correct answer is 1/3. The answer that was marked correct at the time was 1/5.
MANSW took 10 years to change their official answer to 1/3.
Despite the change, many teachers refuse to teach it correctly.

The people who get 1/5 are answering the wrong question ... that answer IS correct for a slightly modified version of the question.
As it stands, the question is VERY easy, but some people choose to overthink the question.

But in the end, the question should not have been asked ... it is conditional probability, which is not in the course.

omg this is from my school's past exam, haha. What really? And we use MIF which doesn't' even thoroughly cover the syllabus

Which year's paper it's from do you remember, mate?

 

[braintic]

omg this is from my school's past exam, haha. What really? And we use MIF which doesn't' even thoroughly cover the syllabus

Which year's paper it's from do you remember, mate?

Its from '97.
I believe many (if not all) solution booklets still give the answer as 1/5.

 

[hit patel]

This is an HSC question, and the last part has caused much disagreement.

The correct answer is 1/3. The answer that was marked correct at the time was 1/5.
MANSW took 10 years to change their official answer to 1/3.
Despite the change, many teachers refuse to teach it correctly.
Wait brain
The people who get 1/5 are answering the wrong question ... that answer IS correct for a slightly modified version of the question.
As it stands, the question is VERY easy, but some people choose to overthink the question.

But in the end, the question should not have been asked ... it is conditional probability, which is not in the course.

Wait Brantic why so? I donot understand. He selects two balls right and drops a red ball which is definite and has another red ball which is definite and so i also get 1/5? can you explain to us why this is so?

 

[braintic]

Wait Brantic why so? I donot understand. He selects two balls right and drops a red ball which is definite and has another red ball which is definite and so i also get 1/5? can you explain to us why this is so?

Firstly ... oops .... I didn't realise that the numbers had been changed from the HSC question. So the answer is not 1/3, it is 1/2.
The simple answer is that, after dropping the red, there are 2 reds left out of 4.
(But I'm sure I will have to resort to the non-simple answer when someone challenges that)
The answer you get by interpreting the question in the incorrect way is actually 1/3.
So I'm not sure how you are getting 1/5.

 

[hit patel]

Firstly ... oops .... I didn't realise that the numbers had been changed from the HSC question. So the answer is not 1/3, it is 1/2.
The simple answer is that, after dropping the red, there are 2 reds left out of 4.
(But I'm sure I will have to resort to the non-simple answer when someone challenges that)
The answer you get by interpreting the question in the incorrect way is actually 1/3.
So I'm not sure how you are getting 1/5.

Wait dont you consider what would happen if both the balls selected were red and 1 red was dropped. and if one ball selected was red then what the probability would be?

 

[braintic]

Wait dont you consider what would happen if both the balls selected were red and 1 red was dropped. and if one ball selected was red then what the probability would be?

If you really need to take cases:

Label the reds R1, R2, R3 and the whites W1, W2

List out all the possible pairs, accounting for order.
If the balls didn't leave your hand, there would be no need to account for order.
BUT, there is a difference between, for example, R1W2 and W2R1, because one of those refers to the red being dropped, while the other refers to the white being dropped. So you must decide that (say) the second item in a pair refers to the ball lying on the floor.

You should get 5 times 4 = 20 different pairs.
This is made up of:
8 pairs with a white on the floor (which must be excluded from the sample space)
6 pairs with a red on the floor but NOT in your hand
6 pairs with a red on the floor AND a red in your hand

As I said, the first 8 are excluded from the sample space because you are told that there is a red on the floor.
So the probability is 6/(6+6) = 1/2



The people who do this incorrectly only exclude from the sample space the pairs that have two whites. This does not account for all the possibilities where a white can end up on the floor. They are answering a different question:

"I tell you (truthfully) that there is at least one red in my hand. What is the probability that my hand in fact holds two reds".

Then the answer is 1/3.