Question

(1 point)
three people get into an elevator on the main floor of a building that has 6 floors (this building has 5 floors above the ground - level). assuming that each of the three people will get off the elevator on one of floors 2 through 6, and each does so randomly.
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part (a) what is the probability that the three people get off on the 4th floor? (use four decimals)
part (b) what is the probability that the three people get off on the same floor? (use four decimals)
part (c) what is the probability that at least two of the three people get off on the same floor? (use four decimals)
part (d) what is the probability that exactly two of the three people get off on the same floor? (use four decimals)

Explanation:

Step1: Calculate total number of outcomes

Each person has 5 choices of floors (floors 2 - 6). So the total number of possible outcomes for 3 people is $n = 5\times5\times5=125$.

Step2: Calculate probability for part (a)

The event that all three people get off on the 4th - floor is 1 outcome. So the probability $P(a)=\frac{1}{125}= 0.0080$.

Step3: Calculate probability for part (b)

There are 5 floors where all three people can get off together. So the probability $P(b)=\frac{5}{125}=0.0400$.

Step4: Calculate probability for part (c)

First, find the probability that all three people get off on different floors. The number of ways for 3 people to get off on different floors is $A_{5}^3=\frac{5!}{(5 - 3)!}=5\times4\times3 = 60$. The probability that all three people get off on different floors is $P(\text{different})=\frac{60}{125}=0.4800$. Then the probability that at - least two people get off on the same floor is $P(c)=1 - P(\text{different})=1-0.4800 = 0.5200$.

Step5: Calculate probability for part (d)

The number of ways to choose 2 out of 3 people is $C_{3}^2=\frac{3!}{2!(3 - 2)!}=3$. The number of ways to choose a floor for these 2 people is 5, and the number of ways to choose a different floor for the remaining 1 person is 4. So the number of favorable outcomes for exactly 2 people getting off on the same floor is $m = C_{3}^2\times5\times4=3\times5\times4 = 60$. The probability $P(d)=\frac{60}{125}=0.4800$.

Answer:

Part (a): 0.0080
Part (b): 0.0400
Part (c): 0.5200
Part (d): 0.4800