Questions: Assume the geometric distribution applies. Use the given probability of success p to find the indicated probability. Find P(3) when p=0.40. P(3)= (Round to five decimal places as needed.)

Assume the geometric distribution applies. Use the given probability of success p to find the indicated probability.

Find P(3) when p=0.40. P(3)= (Round to five decimal places as needed.)

Transcript text: Assume the geometric distribution applies. Use the given probability of success $p$ to find the indicated probability. Find $\mathrm{P}(3)$ when $\mathrm{p}=0.40$. $P(3)=$ $\square$ (Round to five decimal places as needed.)

Solution

Solution Steps

To find the probability of the first success on the third trial in a geometric distribution, we use the formula $ P(X = k) = (1-p)^{k-1} \times p $, where $ k $ is the trial number and $ p $ is the probability of success on each trial. Here, $ k = 3 $ and $ p = 0.40 $.

Step 1: Define the Problem

We need to find the probability $ P(3) $ in a geometric distribution where the probability of success $ p = 0.40 $. The formula for the probability of the first success on the $ k $-th trial is given by:

\[ P(X = k) = (1 - p)^{k - 1} \times p \]

Step 2: Substitute Values

Substituting $ p = 0.40 $ and $ k = 3 $ into the formula, we have:

\[ P(3) = (1 - 0.40)^{3 - 1} \times 0.40 \]

Step 3: Calculate the Probability

Calculating the components:

\[ P(3) = (0.60)^{2} \times 0.40 \]

Calculating $ (0.60)^{2} $:

\[ (0.60)^{2} = 0.36 \]

Now, substituting back:

\[ P(3) = 0.36 \times 0.40 = 0.144 \]