Question

a simulation
using the geometric probability distribution after 500 trials, determine the following probabilities. type your answer in the blank space below.

1. ( p(x = 1) = 0.186 )
2. what is ( p(x > 1) )?

0.997
the probability that it took more than 1 roll would be the sum of the probability of all rolls greater than 1.
(chart with x and p(x) values for x from 1 to ≥30, and number of trials = 500)

Explanation:

Step1: Recall the complement rule

For any event \( A \), \( P(A^c)=1 - P(A) \), where \( A^c \) is the complement of \( A \). Here, the event \( X>1 \) is the complement of \( X = 1 \). So we can use \( P(X>1)=1 - P(X = 1) \).

Step2: Substitute the given value

We know that \( P(X = 1)=0.186 \). So we calculate \( 1-0.186 \).

\( 1 - 0.186=0.814 \)

Wait, but there is a value 0.997 shown. Wait, maybe we need to sum all probabilities where \( X>1 \). Let's check the table.

First, list all \( P(X) \) for \( X = 2,3,\cdots \).

From the table:

\( X = 2 \): \( 0.154 \)

\( X = 3 \): \( 0.146 \)

\( X = 4 \): \( 0.092 \)

\( X = 5 \): \( 0.084 \)

\( X = 6 \): \( 0.056 \)

\( X = 7 \): \( 0.032 \)

\( X = 8 \): \( 0.046 \)

\( X = 9 \): \( 0.024 \)

\( X = 10 \): \( 0.034 \)

\( X = 11 \): \( 0.036 \)

\( X = 12 \): \( 0.026 \)

\( X = 13 \): \( 0.014 \)

\( X = 14 \): \( 0.012 \)

\( X = 15 \): \( 0.016 \)

\( X = 16 \): \( 0.008 \)

\( X = 17 \): \( 0.002 \)

\( X = 18 \): \( 0.004 \)

\( X = 19 \): \( 0.004 \)

\( X = 20 \): \( 0.004 \)

\( X = 21 \): \( 0 \)

\( X = 22 \): \( 0.004 \)

\( X = 23 \): \( 0.002 \)

\( X = 24 \): \( 0 \)

\( X = 25 \): \( 0.004 \)

\( X = 26 \): \( 0 \)

\( X = 27 \): \( 0.006 \)

\( X = 28 \): \( 0 \)

\( X = 29 \): \( 0.002 \)

\( X = 30 \): \( 0.002 \)

Now sum all these:

Let's add step by step:

Start with \( X=2 \) to \( X = 10 \):

\( 0.154+0.146 = 0.3 \); \( 0.3+0.092=0.392 \); \( 0.392 + 0.084=0.476 \); \( 0.476+0.056 = 0.532 \); \( 0.532+0.032 = 0.564 \); \( 0.564+0.046 = 0.61 \); \( 0.61+0.024 = 0.634 \); \( 0.634+0.034 = 0.668 \)

\( X = 11 \): \( 0.668+0.036 = 0.704 \)

\( X = 12 \): \( 0.704+0.026 = 0.73 \)

\( X = 13 \): \( 0.73+0.014 = 0.744 \)

\( X = 14 \): \( 0.744+0.012 = 0.756 \)

\( X = 15 \): \( 0.756+0.016 = 0.772 \)

\( X = 16 \): \( 0.772+0.008 = 0.78 \)

\( X = 17 \): \( 0.78+0.002 = 0.782 \)

\( X = 18 \): \( 0.782+0.004 = 0.786 \)

\( X = 19 \): \( 0.786+0.004 = 0.79 \)

\( X = 20 \): \( 0.79+0.004 = 0.794 \)

\( X = 22 \): \( 0.794+0.004 = 0.798 \)

\( X = 23 \): \( 0.798+0.002 = 0.8 \)

\( X = 25 \): \( 0.8+0.004 = 0.804 \)

\( X = 27 \): \( 0.804+0.006 = 0.81 \)

\( X = 29 \): \( 0.81+0.002 = 0.812 \)

\( X = 30 \): \( 0.812+0.002 = 0.814 \)

Wait, but the initial hint has 0.997. Wait, maybe I misread the problem. Wait, the problem says "after 500 trials", maybe the total probability should sum to 1. Let's check \( P(X = 1)+ \) sum of \( P(X>1) \) should be 1.

\( P(X = 1)=0.186 \), so sum of \( P(X>1) \) should be \( 1 - 0.186=0.814 \). But the 0.997 is maybe a typo or wrong. Wait, but let's re - check the table.

Wait, maybe the table has more values. Wait, the first part of the table:

First table (X from 1 - 10):

X:1 (0.186), 2(0.154), 3(0.146), 4(0.092), 5(0.084), 6(0.056), 7(0.032), 8(0.046), 9(0.024), 10(0.034)

Sum of these: \( 0.186+0.154 = 0.34 \); \( 0.34+0.146 = 0.486 \); \( 0.486+0.092 = 0.578 \); \( 0.578+0.084 = 0.662 \); \( 0.662+0.056 = 0.718 \); \( 0.718+0.032 = 0.75 \); \( 0.75+0.046 = 0.796 \); \( 0.796+0.024 = 0.82 \); \( 0.82+0.034 = 0.854 \)

Wait, no, 0.186 (X=1) + 0.154 (X=2)=0.34; +0.146 (X=3)=0.486; +0.092 (X=4)=0.578; +0.084 (X=5)=0.662; +0.056 (X=6)=0.718; +0.032 (X=7)=0.75; +0.046 (X=8)=0.796; +0.024 (X=9)=0.82; +0.034 (X=10)=0.854.

Then X=11: 0.036, sum=0.854 + 0.036=0.89

X=12: 0.026, sum=0.89+0.026=0.916

X=13: 0.014, sum=0.916+0.014=0.93

X=14: 0.012, sum=0.93+0.012=0.942

X=15: 0.016, sum=0.942+0.016=0.958

X=16: 0.008, sum=0.958+0.008=0.966

X=17: 0.002, sum=0.966+0.002=0.968

X=18: 0.004, sum=0.968+0.004=0.972

X=19: 0.004, sum=0.972+0.004=0.976

X=20: 0.004, sum=0.976+0.004=0.98

X=22: 0.004, sum=0.98+0.004=0.984

X=23: 0.002, sum=0.984+0.002=0.986

X=25: 0.004, sum=0.986+0.004=0.99

X=27: 0.006, sum=0.99+0.006=0.996

X=29: 0.002, sum=0.996+0.002=0.998

X=30: 0.002, sum=0.998+0.002=1.0

Ah, I missed some X values. Let's list all X from 2 to 30:

X=2: 0.154

X=3: 0.146

X=4: 0.092

X=5: 0.084

X=6: 0.056

X=7: 0.032

X=8: 0.046

X=9: 0.024

X=10: 0.034

X=11: 0.036

X=12: 0.026

X=13: 0.014

X=14: 0.012

X=15: 0.016

X=16: 0.008

X=17: 0.002

X=18: 0.004

X=19: 0.004

X=20: 0.004

X=21: 0

X=22: 0.004

X=23: 0.002

X=24: 0

X=25: 0.004

X=26: 0

X=27: 0.006

X=28: 0

X=29: 0.002

X=30: 0.002

Now sum all these:

First, sum X=2 to X=10:

0.154 + 0.146 = 0.3; +0.092 = 0.392; +0.084 = 0.476; +0.056 = 0.532; +0.032 = 0.564; +0.046 = 0.61; +0.024 = 0.634; +0.034 = 0.668

X=11: 0.668 + 0.036 = 0.704

X=12: 0.704 + 0.026 = 0.73

X=13: 0.73 + 0.014 = 0.744

X=14: 0.744 + 0.012 = 0.756

X=15: 0.756 + 0.016 = 0.772

X=16: 0.772 + 0.008 = 0.78

X=17: 0.78 + 0.002 = 0.782

X=18: 0.782 + 0.004 = 0.786

X=19: 0.786 + 0.004 = 0.79

X=20: 0.79 + 0.004 = 0.794

X=22: 0.794 + 0.004 = 0.798

X=23: 0.798 + 0.002 = 0.8

X=25: 0.8 + 0.004 = 0.804

X=27: 0.804 + 0.006 = 0.81

X=29: 0.81 + 0.002 = 0.812

X=30: 0.812 + 0.002 = 0.814

Wait, this is the same as before. But the total probability should be 1. So \( P(X = 1)+ \) sum of \( P(X>1)=0.186 + 0.814 = 1 \), which is correct. So the correct value of \( P(X>1) \) is \( 0.814 \).

Answer:

\( 0.814 \)