Probability of exactly one when events are dependent

[BrowncoatsRule]

Probability of "exactly one" when events are dependent

The question is this:
"An urn is filled with 8 green balls, 2 red balls, and 6 orange balls. Three balls are selected without replacement."

What is the probability that exactly one ball is orange?

I know I could just use the binomial formula if each event were independent (i.e. three balls were selected [with replacement). But I'm not sure how to find the probability in this case because they are dependent events, and the order in which the orange ball is picked affects the probability. I can see three different scenarios of one orange ball being picked:

P(orange picked)*P(not orange picked)*P(not orange picked)
or
P(not orange picked)*P(orange picked)*P(not orange picked)
or
P(not orange picked)*P(not orange picked)*P(orange picked)

The first case corresponds to:
(6/16)(10/15)(9/14)= 9/56
The second case corresponds to:
(10/16)(6/15)(9/14)= 9/56
The third case corresponds to:
(10/16)(9/15)(6/14)= 9/56

Would the probability that exactly one ball is orange just be 3(9/56), or 27/56?

Would a similar process be done if I wanted to know the probability that at most one ball is red?

 

[Bacle2]

Looks right. Yes, a similar methd would work to find out P(at most one is red.)

 

[BrowncoatsRule]

Thanks! And so for the probability that at most one red ball is picked:

"at most one" means either 0 red balls are picked or 1 red ball is picked.

If no red balls are picked, then:
P(not red)*P(not red)*P(not red)= (14/16)(13/15)(12/14)= 13/20

and if one red ball is picked it can happen three different ways:
1) P(red)*P(not red)*P(not red)= (2/16)(14/15)(13/14)= 13/120
2) P(not red)*P(red)*P(not red)= (14/16)(2/15)(13/14)= 13/120
3) P(not red)*P(not red)*P(red)= (14/16)(13/15)(2/14)= 13/120
or P(one red)=3(13/120)= 13/40

So would the probability of at most one red ball be (13/20)(13/40)= 169/800 ?

 

[Bacle2]

I don't see why you're multiplying the two probabilities and not adding them, since the events are disjoint --and disjoint events cannot be independent. Sounds counterintuitive, but it's true.

One way of double-checking your work is by adding P(0 Red)+P(1 Red)+P(2 Reds) and checking that they
add to 1. So try calculating P( exactly 2 Reds ), and you should get 1/40 .

 

[blue_raver22]

I would approach this problem by first considering the definition of dependent events. In this case, the events of selecting balls from the urn without replacement are dependent because the probability of selecting a certain color changes with each selection.

To find the probability of exactly one orange ball being picked, we can use the formula P(A and B) = P(A) * P(B|A), where A and B are two dependent events. In this case, A represents the event of selecting an orange ball and B represents the event of not selecting an orange ball in the next two selections.

So, the probability of exactly one orange ball being picked can be calculated as follows:

P(exactly one orange) = P(orange picked) * P(not orange picked on the next two selections)

= (6/16) * (10/15 * 9/14 + 9/15 * 10/14 + 10/15 * 9/14)

= (3/8) * (1/2 + 3/14 + 3/14)

= (3/8) * (13/14)

= 39/112

Therefore, the probability of exactly one orange ball being picked is 39/112 or approximately 0.348.

To find the probability of at most one red ball being picked, we can use a similar approach. The event of selecting a red ball is still dependent on the previous selections. So, the probability can be calculated as follows:

P(at most one red) = P(no red picked) + P(one red picked)

= P(not red picked)^3 + P(red picked) * P(not red picked on the next two selections)

= (14/16)^3 + (2/16) * (14/15 * 13/14 + 13/15 * 14/14 + 14/15 * 13/14)

= (7/8)^3 + (1/8) * (13/15 + 13/15 + 13/15)

= (7/8)^3 + 13/120

= 343/512 + 13/120

= 399/512

Therefore, the probability of at most one red ball being picked is 399/512 or approximately 0.781.

In summary, when events are dependent, we need to take into account the changing probabilities with each selection in order to accurately calculate the