4. If P(E)= 0.55, P(E or F)=0.80, and P(E and F)=0.05, find P(F). P(F)= _?__ (Si

Question

4. If P(E)= 0.55, P(E or F)=0.80, and P(E and F)=0.05, find P(F).

P(F)= _?__ (Simplify your answer)

5. A golf ball is selected at random from a golf bag. If the golf bag contains 4 green balls, 8 red balls, and 9 black balls, find the probability of the following event.

The golf ball is green or red.

The probability that the golf ball is green or red is ___?__

2. A test to determine whether a certain antibody is present is 99.7% effective. This means that the test will accurately come back negative if the antibody is not present (in the test subject) 99.7% of the time. The probability of a test coming back positive when the antibody is not present (a false positive) is 0.003. Suppose the test is given to seven

randomly selected people who do not have the antibody.

(a) What is the probability that the test comes back negative for all seven people?

(b) What is the probability that the test comes back positive for at least one of the seven

people?

(a) P (all 7 tests are negative)= ___?___ (Round to four decimal places as needed.)

(b)

4. A computer can be classified as either cutting-edge or ancient. 83% of computers are classified as ancient. (a) Two computers are chosen at random. What is the probability that both computers are ancient? (Round to four decimal places as needed.)

(b) Nine computers are chosen at random. What is the probability that all nine computers are ancient? (Round to four decimal places as needed.)

(c) What is the probability that at least one of nine randomly selected computers

is cutting-edge? Would it be unusual that at least one of nine randomly selected computers is cutting edge ? (Round to four decimal places as needed.)

Explanation / Answer

We are allowed to do 1 question at a time. Post again for second question.

4) a) 83% are ancient

P(x = 2) = 0.83 * 0.83 = 0.6889

b) P(x = 9) = 0.83^9 = 0.1869

c) cutting edge = 0.17

P(at least 1 cutting edge) = 1 - (no cutting edge)

P = 1 - 0.83^9

P = 0.8131

No, not unsual

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