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Solution: We can see from the probability distribution, that if a hurricane hits the U.S. coast, there is a .403 probability that it is a category 1 hurricane. a) For permutations, we will use: b) For combinations, we will use: For passwords, order matters. For example, the password 63pn2 is different than the password p2n36 even though both passwords use the exact same characters. Therefore, we want to count the total number of ways that we can choose and order the characters. That means that we should use permutations. We have a total number of nine characters to choose from, so n=9. We will be selecting five characters, so r=5. We will use: Here order does not matter. So, we will use combinations. There are ten countries to choose from, so n=10. Four countries will be chosen, so r=4. We will use
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We are given: We are given: We are given:
In a manufacturing plant, three machines A, B, and C produce 40 %, 35 %, and 25 %, respectively, of the total parts production. The company's quality control department determined that 1 % of the parts produced by machine A, 1.5 % of the parts produced by machine B, and 2 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine B? If we use Def to designate “defective”. We are told that given that a part was produced by machine A, the probability that it has a
defect is: P( Def | A) = .01.
We are told that given that a part was produced by machine B, the probability that it has a
defect is: P( Def | B) =.015.
We are told that given that a part was produced by machine C, the probability that it has a
defect is: P( Def | C) =.02.
Furthermore, we are told that the probability that a part was produced by machine A, B, and C, are respectively:
P( A) = .40, P( B) = .35, P ( C) = .25.
We want to find P ( B | Def), so use:
Suppose A and B are two events with probabilities: P(A)=.35,P(Bc )=.45,P(A∩B)=.25. Find the following: a) P(A∪B). b) P(Ac ). c) P(B). For P(A∪B). Use P(A∪B)=P(A)+P(B)-P(A∩B). But for this equation, we need P(B) which we can find by using P(B)=1-P(Bc ). So, P(B)=1-.45= .55. P(A∪B)=.35+.55-.25=. b. For P(Ac ). Use P(A)=1-P(Ac ) which may be rearranged to (Ac )=1-P(A). P(Ac )=1-.35=.65. c. For P(B). Use (B)=1-P(Bc ). P(B)=1-.45=.55. The probability that a certain type of battery in a smoke alarm will last 3 years or more is .70. The probability that a battery will last 6 years or more is .25. Suppose that the battery is 3 years old and is still working, what is the probability that the battery will last at least 6 years? Define E to be the event that the battery will last 3 years or more. Define F to be the event that the battery will last 6 years or more. We are given that: P(E)=.70. P(F)=.25. Note also that if a battery lasts more than 6 years, it would have had to have lasted 3 years, so P(E∩F)=.25. The question is asking “given event E has occurred, what is the probability that event F will occur”? This may be found by using:
