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

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Person A has a greater chance of picking the Red ball compared to Person B due to having two attempts. The probability for Person A is calculated as 2/5, while Person B's probability is 1/4. This demonstrates that having two attempts increases the likelihood of success. The calculations confirm that Person A's probability of drawing the Red ball is indeed higher than Person B's. Therefore, Person A is more likely to successfully pick the Red ball.
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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.

Help please...
 
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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.

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

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

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:

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

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

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

So, we see for person A we have:

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

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

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

Now, we know:

$$8>5$$

Divide through by 20:

$$\frac{2}{5}>\frac{1}{4}$$

Hence:

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