Jamie has to pull a marble out of a bag containing 3 marbles, one yellow, one blue, and one red. If Jamie were to pull a marble out of the bag, look at its color, and then put it back, what are the chances they would draw blue 3 times in a row?

Answer 1

To determine the probability of Jamie drawing a blue marble three times in a row, we start by calculating the probability of drawing a blue marble on a single draw.

There are three marbles in total: one yellow, one blue, and one red. Thus, the probability of drawing a blue marble on one attempt is:

\[ P(\text{blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{1}{3}. \]

Since Jamie replaces the marble back into the bag after each draw, the situation remains the same for each of the three draws. Therefore, the probability of drawing a blue marble three times in a row is calculated by multiplying the probability of drawing a blue marble for each of the three draws:

\[ P(\text{blue 3 times}) = P(\text{blue}) \times P(\text{blue}) \times P(\text{blue}) = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right). \]

This simplifies to:

\[ P(\text{blue 3 times}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}. \]

Thus, the chances of Jamie drawing a blue marble three times in a row is \(\frac{1}{27}\).

Answer 2

So if Jamie put two more marbles into their bag, one purple and one green, the chances of them getting blue 5 times in a row would be 1/5^5?

Answer 3

Let's first clarify the situation now that Jamie has added two more marbles: one purple and one green. The new total of marbles in the bag is five: one yellow, one blue, one red, one purple, and one green.

Now, the probability of drawing a blue marble in a single attempt is:

\[ P(\text{blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{1}{5}. \]

If Jamie wants to find the probability of drawing a blue marble five times in a row, we multiply the probability of drawing blue for each of the five draws:

\[ P(\text{blue 5 times}) = P(\text{blue}) \times P(\text{blue}) \times P(\text{blue}) \times P(\text{blue}) \times P(\text{blue}) = \left(\frac{1}{5}\right) \times \left(\frac{1}{5}\right) \times \left(\frac{1}{5}\right) \times \left(\frac{1}{5}\right) \times \left(\frac{1}{5}\right). \]

This simplifies to:

\[ P(\text{blue 5 times}) = \left(\frac{1}{5}\right)^5 = \frac{1}{3125}. \]

So the probability of drawing a blue marble five times in a row is \(\frac{1}{3125}\), not \(\frac{1}{5^5}\) as you wrote, but it is indeed equal to \(\frac{1}{5^5}\).