A bag contains 6 large marbles: 3 red, 2 green, and 1 yellow, and 3 sm

MMH91



Ok, first thing you have to understand is that you have to stop going by gut feeling but by rigorous definition.

Independent events are DEFINED as:

P(A n B)=P(A)*P(B).

Now, lets look at E:

Drawing a large one is 6/9, drawing a small one then is 3/8, so in total, we get (6/9)*(3/8). Now compare this to how independent events are defined, and look at the probability of drawing a large vs small marble, INDEPENDENT of what happened before:

We get 6/9 for large, and 3/9 for small (!)

This is exactly the difference: Picking a large affects the probability and (6/9)*(3/8) != (6/9)*(3/9)=P(AnB)=P(A)*P(B) <-> Two events are independent.

No guarantees

scott from TTP.

can u please provide an explanation for this

Examining B.

Independent if P(AandB)= P(A)*P(B).

P(Small not Green) =2/9
P(Green given above)=3/8
Probability of both: 6/72=1/12

P(Small and Green)= 1/9
P(Green given above)= 2/8
Probability of both:
2/72=1/36

Total probability: 1/12+1/36=
4/36= 1/9

Checking to see if this equals P(Small)*P(Green):

P(Small)= 3/9=1/3
P(Green)=3/9=1/3

P(Small)*P(Green)= 1/3*1/3=

1/9.

Since the probabilities of 1/9 are the same, the events are independent

Another approach:

If P(B/A)= P(B) then independent.

B is small, A green.

Four possibilities:

Pick small green, then green:

1/9*2/8 =1/36

Pick large green, then green:

2/9*2/8= 2/36

Pick Small not Green, then Green:

2/9*3/8= 3/36

Pick Large not Green, then Green:

4/9*3/8 = 6/36

Probability of Small given Green is then (eliminating common denominator of 36):

(1+3)/(1+3+2+6)= 4/12=

1/3.

So, if this is equal to the a priori probability of a small marble, then the events are independent:

Probability of a small marble:

3/9= 1/3

Probabilities are the same so the events are independent


Posted from my mobile device

One definition for independence is the following: events A and B are said to be independent if P(A) = P(A | B). Here, P(A | B) is the probability of A given B, meaning the probability of A assuming event B happened. Using this definition, we can analyze each answer choice:
A) Event A is "selecting a large marble" and event B is "selecting a red marble". Thus, P(A) = 6/9 = 2/3. If we assume event B already happened, i.e. if we assume the selected marble is red, then the probability that it is a large marble is 3/4 (since there are 4 red marbles in total and 3 of them are large). Since P(A) ≠ P(A | B), events A and B are not independent.

B) Event A is "selecting a small marble" and event B is "selecting a green marble". Thus, P(A) = 3/9 = 1/3. If we assume event B already happened, i.e. if we assume the selected marble is green, then the probability that it is also small is 1/3 (since there are 3 green marbles in total and 1 of them is green). Since P(A) = P(A | B), events A and B are independent. The answer is B.

It has been suggested in earlier comments that the events in answer choice E look like they should be independent. Let's analyze answer choice E as well:

E) Event A is "selecting a large marble" and event B is "selecting a small marble". Here, P(A) = 2/3. However, if we assume that event B already happened, i.e. if the selected marble is small, then the probability that the selected marble is large is 0. In other words, P(A | B) = 0. Since P(A) is not equal to P(A | B), events A and B are not independent. In this answer choice, events A and B are mutually exclusive. It is a good idea to note that two mutually exclusive events are never independent unless one of the events have 0 probability.

Now I'm confused.

The analysis of the other answer choices suggest that there are two draws involved.

The above however discusses large and small occurring in one draw.

I would have said P(A) = 2/3 for large ball.

P(B) = 1/3

P(A|B) is probability of a large ball on second draw given small ball drawn first.

Since there are now 8 balls left, 6 large,2 small,

Probability of large ball is now 3/4= P(A|B)

Since P(A|B)=3/4 is not equal to P(A)= 2/3, the events are not independent.

What am I missing?

Posted from my mobile device

You are right. There are two draws. Your calculations of P(A) and P(A | B) are also correct. However, the fact that P(A) is not equal to P(A | B) actually shows that A and B are not independent.

Ok thanks. Yes missed the "not"

Of course.

I found the explanation of ScottTargetTestPrep the most usefull and clear.

As he mentions, 2 events A and B are said to be independent if P(A)=P(A|B), P(A|B) being the probability of picking A within the universe of B. It is helpfull to visualize it as a matrix