Question 1185647: A fruit grower claims that 2/3 of his mango crop has been contaminated by a fruit fly infestation. Find the probability that of the next four mangoes inspected by this grower, 1 to 3 m inclusive of mangoes are contaminated
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p = 2/3 = probability any mango picked at random is contaminated.
q = 1 - 2/3 = 1/3 = probability any mango picked at random is not contaminated.
binomial probability states that:
p(x) = p^x * q^(n-x) * c(n,x)
if n = 4, then the total probabilities will be:
p(0) + p(1) + p(2) + p(3) + p(4).
p(1) = (2/3)^1 * (1/3)^(4-1 = 3) * c(4,1) = .098765432
p(2) = (2/3)^2 * (1/3)^(4-2 = 2) * c(4,2) = .296296296
p(3) = (2/3)^3 * (1/3)^(4-3 = 1) * c(4,3) = .395061728
add them up and the probability of 1 to 3 rotten fruits out of 4 = .790123457.
the total probabilities are equal to 1 as shown below.
p(0) = 0.012345679
p(1) = 0.098765432
p(2) = 0.296296296
p(3) = 0.395061728
p(4) = 0.197530864
p(total) = 1
p(1,2,3) = 0.790123457
c(n,x) is the number of ways you can get x elements out of a set of n.
the formula is c(n,x) = n! / (x! * (n-x)!)
for example:
c(4,3) is equal to 4! / (3! * 1!) = (4 * 3!) / (3! * 1!) = 4 / 1 = 4.
1! is always equal to 1
0! is always equal to 1 as well.
as an example, p(0) = (2/3)^0 * (1/3)^4 * c(4,0) = 1 * .012345679 * 1 = .012345679.
your solution is that the probability of getting 1 to 3 contaminated mangoes is equal to .790123457.
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