To find the probability that at least one red gumball is drawn in 5 drawings, with replacement, follow these steps:

Step 1: Total Probability

The total number of gumballs in the bag is: $3 + 4 + 2 = 9$

The probability of drawing a red gumball is: $P(\text{Red}) = \frac{4}{9}$

The probability of not drawing a red gumball in one draw is: $P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - \frac{4}{9} = \frac{5}{9}$

Step 2: Probability of No Red Gumballs in 5 Draws

The probability of not drawing a red gumball in 5 consecutive draws is: $P(\text{No Red in 5 Draws}) = \left(\frac{5}{9}\right)^5$

Calculate this value: $P(\text{No Red in 5 Draws}) = \left(\frac{5}{9}\right)^5 = \frac{3125}{59049} \approx 0.0529$

Step 3: Probability of At Least One Red Gumball

The probability of at least one red gumball is the complement of not drawing any red gumballs: $P(\text{At Least One Red}) = 1 - P(\text{No Red in 5 Draws})$

Substitute the value: $P(\text{At Least One Red}) = 1 - 0.0529 \approx 0.9471$

Final Answer:

$P(\text{At Least One Red}) \approx 0.95$

Rounded Answer:

0.95

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Would you like more detailed steps or have any questions?
Here are 5 related questions to expand your understanding:

1. How would the probability change if the gumballs were not replaced after each draw?
2. What is the probability of drawing exactly 2 red gumballs in 5 draws with replacement?
3. How does the probability of not drawing any red gumballs change with more draws?
4. If a new color of gumballs is added, how do the probabilities adjust?
5. Can you calculate the expected number of red gumballs drawn in 5 trials?

Tip: Always consider whether the events are independent (like here, with replacement) or dependent, as it significantly affects the probability calculations!