Question

There are 17 blue, 5 green, and 3 red balls in a jar. We randomly select a ball and return it back to the jar. We repeat it 8 times. To find the probability of the event that every time we get a green ball we need to use the following distribution:
Group of answer choices

binomial with number of trials 8 and probability of success 0.8

Poisson Distribution with average 5

uniform distribution

binomial with number of trials 8 and probability of success 0.2

binomial with number of trials 5 and probability of success 0.2

binomial with number of trials 25 and probability of success 0.2

Poisson Distribution with average 8

binomial with number of trials 25 and probability of success 0.5

binomial with number of trials 8 and probability of success 1/3

Answer 1

To determine the correct distribution to describe the situation where we are randomly selecting a ball from a jar containing blue, green, and red balls, we first need to identify the total number of balls in the jar and the probability of success (selecting a green ball).

**Given:**
- Number of blue balls = 17
- Number of green balls = 5
- Number of red balls = 3

**Total number of balls:**
\[
17 \text{ (blue)} + 5 \text{ (green)} + 3 \text{ (red)} = 25 \text{ (total)}
\]

**Probability of selecting a green ball (success):**
\[
P(\text{green}) = \frac{\text{Number of green balls}}{\text{Total number of balls}} = \frac{5}{25} = \frac{1}{5} = 0.2
\]

Since the ball is returned to the jar after each selection, the trials are independent, making this a scenario suitable for a binomial distribution.

**Parameters:**
- Number of trials (n) = 8 (since we are drawing 8 times)
- Probability of success (p) = 0.2 (the probability of selecting a green ball)

Thus, the situation can be modeled by:

**Binomial distribution with number of trials = 8 and probability of success = 0.2.**

So, the correct choice is:
- **binomial with number of trials 8 and probability of success 0.2**