Question 1137553: Naval intelligence reports that 6 enemy vessels in a fleet of 21 are carrying nuclear weapons. If 8 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 Theo(13342)   (Show Source):
You can put this solution on YOUR website! the probability that more than 1 of the vessels targeted carried nuclear weapons would be 1 - p(0) - p(1).
let p = the probability that any one boat targeted carried nuclear weapons.
let q = the probability that any one boat targeted did not carry nuclear weapons.
p = 6/21.
q = 1 - p = 1 - 6/21 = 21/21 - 6/21 = 15/21.
q + p is always equal to 1.
formula for p(x) is p(x) = p^x * q^(n-x) * c(h,x)
x is the probability of getting x elements out of n .
n is the number of possible elements.
in this problem, x will be from 0 to 8 and n will be 8.
p(0) is therefore p^0 * q^(8-0) * c(8,0) = p^0 * q^8 * c(8,0)
p(1) is therefore p^1 * q^(8-1) * c(8,1) - p^1 * q^7 * c(8,1)
since p = 6/21 and q = 15/21, the formulas become:
p(0) = (6/21)^0 * (15/21)^8 * c(8,0)
p(1) = (6/21)^1 * (15/21)^7 * c(8,1)
the formula for c(n,x) is equal to n! / (x! * (n-x)!).
c(8,0) is equal to 8! / (0! * 8!) which is equal to 8! / 8! which is equal to 1.
0! is always equal to 1.
c(8,1) is equal to 8! / (1! * 7!) which is equal to (8 * 7!) / (1! * 7!) which is equal to 8/1 * 7! / 7! which is equal to 8.
p(0) is therefore equal to (6/21)^0 * (15/21)^8 * 1 = .0677603615.
p(1) is therefore equal to (6/21)^1 * (15/21)^7 * 8 = .2168331569.
p(>1) is equal to p(2) + ..... + p(8), which is equal to 1 - p(0) - p(1) which is equal to .7154064815.
the sum of all probabilities must be equal to 1 as shown in the following excel spreadsheet printout.
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