6. Exercise 5.27 METHODS AND APPLICATIONS The January 1986 mission of the Space

Question

6. Exercise 5.27 METHODS AND APPLICATIONS The January 1986 mission of the Space Shuttle Challenger was the 25th such shuttle mission. It was unsuccessful due to an explosion caused by an O-ring seal failure (a) According to NASA, the probability of such a failure in a single mission was 1/59,626. Using this value of p and assuming all missions are independent, calculate the probability of no mission failures in 19 attempts. Then calculate the probability of at least one mission failure in 19 attempts. (Do not round your intermediate calculation and round your final answers to 4 decimal places.) P(x = 0) P(x 2 1) (b) According to a study conducted for the Air Force, the probability of such a failure in a single mission was 1/35. Recalculate the probability of no mission failures in 19 attempts and the probability of at least one mission failure in 19 attempts. (Do not round your intermediate calculation and round youn final answers to 4 decimal places.) P(x = 0) P(x 2 1) (c) Based on your answers to parts a and b, which value of p seems more likely to be true? Explain. (Enter answer as a division like xly.) 1/36 (Click to select) 1159, 626 1/35 (d) How small must p be made in order to ensure that the probability of no mission failures in 19 attempts is .999? (Round your answers to 5 decimal places.)

Explanation / Answer

PROBABILITY OF REMOVING THE STAIN = 0.50

PROBABILITY OF NOT REMOVING STAIN = 0.5

WE HAVE N = 20

THEREFORE THE PROBABILITY P(X<=17) = 1 - [P(18)+P(19)+P(20)]

P(18) = 20C18*(0.5)^20 = 0.0003

P(19) = 20C19*(0.5)^20 = 0.00001

P(20) = 0.5^20 = 0.0000001

THEREFORE P(X<=17) = 1 - 0.0003+0.00001+0.0000001 = 0.99969

B) IF THE X TURNS OUT TO BE REAL IN 17 THEN ACCORDING TO THE PROBABILITY IT IS A STRONG EVIDENCE.

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