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

1) A) PROBABILITY OF FAILURE = 1/59626 = 0.000016

PROBABILITY OF SUCCESS = 0.99998

P(0 FAILURE ) = 19C0*(0.000016)^0(0.99998)^19 = 0.99962

P( AT LEAST 1 FAILURE ) = 1 - 0.99962 = 0.00038

B) PROBABILITY OF FAILURE = 0.028

PROBABILITY OF SUCCESS = 0.972

P(X=0) = 19C0*(0.028)^0*(0.972)^19 = 0.582

P(ATLEAST 1 FAILURE) = 1-0.582 = 0.418

C) ACCORDING TO THE PROBABILITIES IN PART A AND PART B THE MORE CORRECT PROBABILITY = 1/35

D) P(X=0) = 19C0(P)^0*(Q)^19 = 0.999

THEREFORE Q = 0.999^(1/19) = 0.99947

THEREFORE = = 0.00053

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