If a procedure meets all of the conditions of a binomial distribution except the number of trials is not fixed, then the geometric distribution can be used. the probability of getting the first success on the xth trial is given by p(x)equalsp (1 minus p )superscript x minus 1, where p is the probability of success on any one trial. subjects are randomly selected for a health survey. the probability that someone is a universal donor (with group o and type rh negative blood) is 0.05. find the probability that the first subject to be a universal blood donor is the fourth person selected.

The probability is 0.0428

Step-by-step explanation:

First, let's remember that the binomial distribution is given by the formula:

where k is the number of successes in n trials and p is the probability of success.

However, the problem tells us that when there isn't a number of trials fixed, we can use the geometric distribution and the formula for getting the first success on the xth trial becomes:

The problem asks us to find the probability of the first success on the 4th trial (given that the first subject to be a universal blood donor will be the fourth person selected)

Using this formula with the parameters given, we have:

p = 0.05

x = 4

Substituting these parameters in the formula and solving it, we get:

Therefore, the probability that the first subject to be a universal blood donor is the fourth person selected is 0.0428 or 4.28%

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

sorry, i can't understand your question. will later if you describe more clearly!

: )