SOLUTION: Naval intelligence reports that 7 enemy vessels in a fleet of 21 are carrying nuclear weapons. If 6 vessels are randomly targeted and destroyed, what is the probability that more t

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Question 1148872: Naval intelligence reports that 7 enemy vessels in a fleet of 21 are carrying nuclear weapons. If 6 vessels are randomly targeted and destroyed, what is the probability that more than 1 vessel transporting nuclear weapons was destroyed? Express your answer as a fraction or a decimal number rounded to four decimal places.
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
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Probability  P(more than 1 vessel) = P(2 vessels) + P(3 vessels) + P(4 vessels) + P(5 vessels) + P(6 vessels).


P(2 vessels) = %287%2F21%29%2A%286%2F20%29                 =  0.1


P(3 vessels) = %287%2F21%29%2A%286%2F20%29%2A%285%2F19%29             = 0.026316


P(4 vessels) = %287%2F21%29%2A%286%2F20%29%2A%285%2F19%29%2A%284%2F18%29         = 0.005848


P(5 vessels) = %287%2F21%29%2A%286%2F20%29%2A%285%2F19%29%2A%284%2F18%29%2A%283%2F17%29     = 0.001032


P(6 vessels) = %287%2F21%29%2A%286%2F20%29%2A%285%2F19%29%2A%284%2F18%29%2A%283%2F17%29%2A%282%2F16%29 = 0.000129


Probability  P(more than 1 vessel) is the sum of the numbers in the last column, which is  0.133325  (approximately),

or   0.1333,  rounded as requested.

Solved.