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At the World Championship Dart contest, bets were being placed on the accuracy of the contestants....

At the World Championship Dart contest, bets were being placed on the accuracy of the contestants. One of the braggarts said his accuracy was only 80% but he wanted to play with the “Big Boys”. So they gave him three darts and the object of the betting contest was to see how many of these three darts he could stuff into the Bull’s Eye. So if his accuracy was only 80%, then determine from first principles (not using the Binominal chart or the Binominal formula but using the Union, Complement, Intersection and Combination rules) what would be the betting line (probabilities) for the following scenarios.

  1. What is the probability that he could get all three darts in the Bull’s eye? (1)
  2. What is the probability that he could get two of the three darts in the Bull’s eye? (1)
  3. What is the probability that he could get one of the three darts in the Bull’s eye? ( 1)
  4. What is the probability that none of his darts found the Bull’s eye? (1)
  5. What is the probability that at least one of his darts found the Bull’s eye? (1)
math   Statistics-And-Probability   
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Given, number of darts that he could stuff into the Bulls eye 3. Let H be the event that dart will hit the Bull eye. Let H be the event that dart will not hit the Bull eye. The probability that dart will hit the bull eye is p(H) 0.80. The probability that dart will not hit the bull eye is p(H)-1-P(H)-1-0.80-0.20 a) The probability that he could get all three darts in the Bulls eye is, P(H)xP(H)xP(H) 0.80x0.80x0.80 0.512 b) The probability that he could get two of the three darts in the Bulls eye is, 3(0.80x 0.80x0.20) -3(0.128) 0.384 The probability that he could get one of the three darts in the Bulls eye is,3(0.80x0.20x 0.20) 3(0.032 0.096 d) The probability that none of his darts found the Bulls eye is, e) Find the probability that at least one of his darts found the Bulls eye We know that, P(At least one of the event occurs) 1- P(None of the event occurs) At least one of his None of his darts darts found the Bulls eye found the Bulls eye = 1-(0.20× 0.20 × 0.20) = 1-0.008 - 0.992

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