Which Person (if any) has a greater chance of picking out the Red ball?

[pwange]

A)
There are 5 balls inside bag A. They're all a different color (Red, Blue, Green, Orange, Purple).

Person A has two attempts to pick a Red ball from inside bag A:

1st Attempt: 1/5 chance = 20% = Let's say Person A picked out a blue ball.

2nd Attempt: 1/4 chance = 25%.

B)
There are 4 balls inside bag B. They're all a different color (Red, Green, Orange, Purple).

Person B have one attempt to pick a Red ball from inside bag B:

1st Attempt: 1/4 chance = 25%.

Question:
Which Person (if any) has a greater chance of picking out a Red ball?

My Answer:
Person A has a higher chance, because they get two attempts. Although I have no clue how to prove that.
Nor do I know by how much.

The answer might be equal (25% on both), but that doesn't make much sense to me.
I'd rather have two attempts than one.

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Help please...

 

[MarkFL]

For person A, we can approach this directly:

We look at the case where red was drawn first, OR red was not drawn first but is drawn second.

\(\displaystyle P(\text{Red})=P(\text{First draw red})+P(\text{First not red AND second red})\)

\(\displaystyle P(\text{Red})=\frac{1}{5}+\frac{4}{5}\cdot\frac{1}{4}=\frac{2}{5}\)

We can also use the complement rule:

We know it is certain the person A will draw red in the first two draws, OR will not draw red in the first two draws:

\(\displaystyle P(\text{Red})+P(\text{Not red})=1\)

\(\displaystyle P(\text{Red})=1-P(\text{Not red})\)

\(\displaystyle P(\text{Red})=1-\frac{4}{5}\cdot\frac{3}{4}=1-\frac{3}{5}=\frac{2}{5}\)

So, we see for person A we have:

\(\displaystyle P_A(\text{Red})=\frac{2}{5}\)

And it is easy to see that for person B we have:

\(\displaystyle P_B(\text{Red})=\frac{1}{4}\)

Now, we know:

\(\displaystyle 8>5\)

Divide through by 20:

\(\displaystyle \frac{2}{5}>\frac{1}{4}\)

Hence:

\(\displaystyle P_A(\text{Red})>P_B(\text{Red})\)