Assume that a procedure yields a geometric distribution where the probability of success is 19%. use the geometric probability formula to find the probability that the first success occurs on the third trial.
a)0.469

b)0.207

c)0.125

d)0.544

The probability that the first success occurs on the third trial is 0.125.

Option C

Step-by-step explanation:

The probability of success is given by the geometric distribution formula:

Where,

p = probability of success for single trial.

q = probability of failure for a single trial (1-p)

x = the number of failures before a success.

P(X = x) = Probability of x successes in n trials.

Given:

p = 19% = 0.19

q = 1 – p = 1 – 0.19 = 0.81

x = 3

We want to find the probability that the first success occurs on the third trial so,

c is the correct answer

step-by-step explanation:

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: