Question

An urn contains 4 colored balls: 2 orange and 2 blue. 2 balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also orange.
Please, guide me step by step.

Answer 1

@SithsAndGiggles

Answer 2

Let \(B_1,B_2,O_1,O_2\) denote the events that the first ball is blue, second ball is blue, first ball is orange, and second ball is orange (respectively).

You want to find \(P\bigg((O_1\cap O_2)~|~(O_1\cup O_2)\bigg)\), correct?

Answer 3

Honestly, I don't get the concept. Please, just give me the guidance. If I don't understand, I will ask.

Answer 4

Have you learned conditional probabilities yet?

Answer 5

Let me arrange your guidance:
\(B_1 \)= {the first ball is blue}\(\rightarrow P (B_1) = \dfrac{1}{4}\)
Yes, I do.

Answer 6

the same with all of them , so that
P(B2) = P (O1) =P(O2) = 1/4, right?

Answer 7

With the additional information that atleast one of the ball is Orange, the sample space shrinks to :
\[\large \{OB, ~BO, ~OO\}\]

Answer 8

@rational but we have to label the set before counting, right?

Answer 9

Not quite.
\[P(B_1)=P(O_1)=\dfrac{1}{4}\]
be this is not the case for the second balls because the balls are not being replaced.

Answer 10

Yes, for the second time, the denominator should be 3. But I don't know how to arrange the sets

Answer 11

\[P\bigg((O_1\cap O_2)~|~(O_1\cup O_2)\bigg)=\frac{P\bigg[(O_1\cap O_2)\cap(O_1\cup O_2)\bigg]}{P\bigg[O_1\cup O_2\bigg]}\]

Answer 12

Probability = (# of favorable outcomes) / (# of total outcomes)

since the sample space has a total of 3 outcomes and 1 is in your favor, isn't probability simply 1/3 ?

Answer 13

only 1 option I can see.