Hey guys, just 3 questions this time...

1. Three different machines M1, M2, and M3 were used for producing a large

batch of similar manufactured items. Suppose that 20% of the items were

produced by M1, 30% by M2, and 50% by M3. Suppose further that 1% of the items produced by M1 are defective, that 2% of the items produced by M2 are defective, and that 3% of the items produced by M3 are defective.

i) Suppose one item is selected at random from the entire batch and it

is found to be defective. Determine the probability that this item

was produced by M2.

ii) Suppose one item is selected at random from the entire batch and it

is found to be nondefective. Determine the probability that it was

produced by M2.

My answers:

We must use conditional probability.

i)

$\displaystyle P( M2 \cap Id) = P(M2) \cdot P(Id | M2)$Code:M1 M2 M3 Items Produced (Ip) : 0,2 0,3 0,5 Items Defective (Id) : 0,01 0,02 0,03 Items not-Defect. (In): 0,99 0,98 0,97

$\displaystyle = 0,3 \times 0,02$

$\displaystyle = 0,006$

$\displaystyle = 0,6$%

ii) P(M2 \cap In) = P(M2) \cdot P(In | M2)

$\displaystyle = 0,3 \times 0,98$

$\displaystyle = 0,294$

$\displaystyle = 29,4$%

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2. A box contains 3 coins with head on each side, 4 coins with a tail on each side, and 2 fair coins. One of these 9 coins is selected at random and tossed once. Suppose a head is obtained. Determine the probability that the coin is one with head on each side.

My answer:

We obtain a head so we know that it is not one of the coins with the tail on each side. So our coin will be one out of the 3 head-only coins and the 2 fair coins.

We have 5 coins and 3 of them have two heads.

So the probability is 3 out of 5 equals 60%.

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3. In a certain city, 30% of the people are Conservatives, 50% are Liberals, and 20% are Independents. Records show that in a particular election, 65% of the Conservatives voted, 82% of the liberals voted, and 50% of the Independents voted. If a person in the city is selected at random and it is learned that he did not vote in the last election, what is the probability that he is a Liberal?

My answer:

We must use conditional probability.

$\displaystyle P(Lib \cap N) = P(Lib) \cdot P(N | Lib)$Code:Consv Lib Ind Population: 0,3 0,5 0,2 Voted(V): 0,65 0,82 0,5 Not Voted(N): 0,35 0,18 0,5

$\displaystyle = 0,5 \times 0,18$

$\displaystyle = 0,09$

$\displaystyle = 9$%