Question

assume the geometric distribution applies. use the given probability of success p to find the indicated probability. find p(4) when p = 0.70. p(4) = \boxed{} (round to five decimal places as needed.)

Explanation:

Step1: Recall Geometric Distribution Formula

The probability mass function for a geometric distribution is \( P(X = k)=(1 - p)^{k - 1}p \), where \( p \) is the probability of success, and \( k \) is the number of trials until the first success. Here, \( p = 0.70 \) and \( k = 4 \).

Step2: Substitute Values into Formula

First, calculate \( 1 - p \): \( 1 - 0.70 = 0.30 \). Then, calculate \( (1 - p)^{k - 1} \): \( (0.30)^{4 - 1}=(0.30)^{3}=0.027 \). Multiply by \( p \): \( 0.027\times0.70 = 0.0189 \). Wait, no, wait: \( k = 4 \), so \( (1 - p)^{4 - 1}=0.3^{3}=0.027 \), then \( 0.027\times0.7 = 0.0189 \)? Wait, no, let's recalculate. \( 0.3^3 = 0.027 \), then \( 0.027\times0.7 = 0.0189 \)? Wait, no, maybe I made a mistake. Wait, \( (1 - p)=0.3 \), \( k - 1 = 3 \), so \( 0.3^3 = 0.027 \), then multiply by \( p = 0.7 \): \( 0.027\times0.7 = 0.0189 \). But wait, let's check the formula again. The geometric distribution can be defined as the number of trials until the first success, so \( P(X = k)=(1 - p)^{k - 1}p \). So for \( k = 4 \), it's \( (0.3)^{3}\times0.7 \). Let's compute \( 0.3^3 = 0.027 \), then \( 0.027\times0.7 = 0.0189 \). But wait, the problem says to round to five decimal places. Wait, 0.0189 is four decimal places. Wait, maybe I miscalculated. Wait, \( 0.3^3 = 0.027 \), \( 0.027\times0.7 = 0.0189 \), which is 0.01890 when rounded to five decimal places? Wait, no, 0.0189 is 0.01890 when rounded to five decimal places. Wait, but let's do it more accurately. \( 0.3\times0.3 = 0.09 \), \( 0.09\times0.3 = 0.027 \), \( 0.027\times0.7 = 0.0189 \). So \( P(4)=0.01890 \) when rounded to five decimal places? Wait, no, 0.0189 is 0.01890. Wait, but maybe I made a mistake in the formula. Wait, another way: the geometric distribution formula is also sometimes defined as the number of failures before the first success, but the problem says "the geometric distribution applies" with probability of success \( p \), so the formula \( P(X = k)=(1 - p)^{k - 1}p \) for \( k = 1,2,3,\dots \) (number of trials until first success). So for \( k = 4 \), it's the probability that the first success occurs on the 4th trial, so three failures and then a success. So probability of failure is \( 1 - p = 0.3 \), three failures: \( 0.3^3 \), then success: \( 0.7 \), so \( 0.3^3\times0.7 = 0.027\times0.7 = 0.0189 \), which is 0.01890 when rounded to five decimal places.

Answer:

0.01890