Whitus v. Georgia In the classic legal case of Whitus v. Georgia, a jury pool of 90 people was supposed to be randomly selected from a population in which 27% were minorities. Among the 90 people selected, 7 were minorities. Find the probability of getting 7 or fewer minorities if the jury pool was randomly selected. Is the result of 7 m

0.000004 = 0.0004% probability of getting 7 or fewer minorities if the jury pool was randomly selected.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either it was a minority, or it was not. The probability of a person being a minority is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

Jury of 90 people:

This means that

27% were minorities.

This means that

Find the probability of getting 7 or fewer minorities if the jury pool was randomly selected.

This is:

In which

So

0.000004 = 0.0004% probability of getting 7 or fewer minorities if the jury pool was randomly selected.

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

2 apples

step-by-step explanation:

taking away equals subtraction. 6-4=2