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If cities A, B, C, and D are visited in random order, each city visited once, find the probability that city D will be visited first, city B second, city A third, and city Clast.

User Nikitas
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Answer: So, the probability is 0.25 or 25%.

Note: The probability assumes that all possible permutations are equally likely.

Step-by-step explanation: To find the probability of visiting cities in a specific order, we need to consider the total number of possible permutations (arrangements) of visiting the cities. In this case, we have 4 cities (A, B, C, D) to visit in random order.

The total number of possible permutations can be calculated using the factorial function. The factorial of a number n, denoted as n!, is the product of all positive integers less than or equal to n.

In this case, we have 4 cities, so the total number of possible permutations is 4! = 4 x 3 x 2 x 1 = 24.

Now, we want to find the probability of visiting the cities in the specific order: D first, B second, A third, and C last.

To calculate this probability, we need to determine the number of favorable outcomes (i.e., the number of permutations that satisfy the given condition). In this case, we want to find the number of permutations where D is visited first, B is visited second, A is visited third, and C is visited last.

Since we want D to be visited first, we fix D in the first position. Now, we have 3 remaining cities (B, A, C) to arrange in the remaining 3 positions.

The number of permutations of 3 cities can be calculated as 3! = 3 x 2 x 1 = 6.

Therefore, the number of favorable outcomes is 6.

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

Probability = Number of favorable outcomes / Total number of possible outcomes

Probability = 6 / 24

Simplifying, we find that the probability of visiting city D first, city B second, city A third, and city C last is 1/4 or 0.25.

User Chris Mack
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