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I am Lyosha [343]
3 years ago
7

1. State if each scenario involves a permutation or a combination. Then find the number of possibilities:(a) Castel and Joe are

planning trips to three countries this year. There are 7 countries they would like to visit. One trip will be one week long, another two days, and the other two weeks.(b) You are setting the combination on a three-digit lock. You want to use the numbers 123 but don't care what order they are in.(c) A group of 25 people are going to run a race. The top 8 finishers advance to the finals.
Mathematics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer:

a) This is a permutation, there are 210 ways to do it

b) This is a permutation with 6 possibilities

c) This is a combination with a total of 1081575 possibilities

Step-by-step explanation:

a) Here the order does matter bacause it is not the same that they go to France in the weak trip than they go there for 2 days, as a result, this scenario is a permutation.

You need to pick 3 countries from a total of 7, but the order matters, so the total amount of ways to do this is

{7 \choose 3} * 3! = \frac{7!}{(7-3)!} = 210

b) Here each permutation of 123 will give you a possible (and a different one from the others) lock-combination, so this is again a permutation.

The total amount of ways we have to do the selectio of the lock-combination is the total amount of permutations of a set of 3 elements, in other words, it is 3! = 6.

c) If you only cares about who advances to the finals, then the order doesnt matter here, you just want 8 people out of the 25 available. This is therefore a combination, and the total amount of possible cases are

{25 \choos 8} = 1081575 .

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Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

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P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

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P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

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