Due in 4 hours, 35 minutes. Due Sun 02/25/2018 11:59 pm without replacement and

Question

Due in 4 hours, 35 minutes. Due Sun 02/25/2018 11:59 pm without replacement and the entire batch is accepted if every With one item in the sample is okay. The ABCD Electronics Company has just manufactured 7000 write-rewrite CDs, and 160 are defective. If 5 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted? Report the answer as a sampling, a sample of items is randomly se [10/10] [10/10 percent rounded to one decimal place accuracy. You need not enter the "%" symbol. 4 (10/10) prob 5 [10/10] 6 [10/10) 7 (0/10) 8 (0/10) Q9 (10/10] Q 10 (10/10) Q 11 [10/10) Q 12 [10/10 Q13 (0/10) Q14 [10/10 Q15 (10/10) Q16 (10/10) Points possible: 10 Q 17 [10/10] Q18 [10/101 C Q19 (0/10) Q 20 [10, 10] Q21 (10/10) Q 22 [10/ 10] Print Version This Mac can't

Explanation / Answer

Answer:

Probability of first one is accepted = (6840/7000)

Probability of first two is accepted = (6840/7000)*(6839/6999)

Probability of first three is accepted = (6840/7000)*(6839/6999)*(6838/6998)

Probability of first four is accepted = (6840/7000)*(6839/6999)*(6838/6998)*(6837/6997)

Probability of five or the lot is accepted = =(6840/7000)*(6839/6999)*(6838/6998)*(6837/6997)*(6836/6996)

=0.890791

Prob = 89.1 %

P( 10 or 9) = 1/12+1/12

= 0.166667

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