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Question
Five numbers x1, x2, x3, x4, x5 are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order such that x1 < x2 < x3 < x4 < x5. What is the probability that x2 = 7 and x4 = 11?
Options
`26/51`
`3/104`
`1/68`
`1/34`
Solution
`bb(1/68)`
Explanation:
To determine the probability that x2 = 7 and x4 = 11 when five numbers are randomly selected from the set {1, 2, 3, ..., 18} and arranged in increasing order, follow these steps:
1. Fix x2 and x4:
- x2 = 7
- x4 = 11
2. Select the remaining three numbers:
- We need one number x1 < 7.
- We need one number 7 < x3 < 11.
- We need one number x5 > 11.
3. Count the possible choices for each number:
- Numbers less than 7: {1, 2, 3, 4, 5, 6} - 6 choices.
- Numbers between 7 and 11: {8, 9, 10} - 3 choices.
- Numbers greater than 11: {12, 13, 14, 15, 16, 17, 18} - 7 choices.
4. Calculate the number of ways to choose these numbers:
- 6 choices for x1.
- 3 choices for x3.
- 7 choices for x5.
- Therefore, the total number of ways to choose x1, x3 and x5 is 6 × 3 × 7 = 126.
5. Total number of ways to choose any 5 numbers from 18:
- The number of ways to choose 5 numbers from 18 is given by the combination formula `(18/5)`.
- `(18/5) = (18!)/(5!(18 - 5)!) = (18!)/(5!*13!)`.
- `(18/5) = (18 xx 17 xx 16 xx 15 xx 14)/(5 xx 4 xx 3 xx 2 xx 1) = 8568`
6. Calculate the probability:
- The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.
- Probability = `"Number of favorable outcomes"/"Total number of possible outcomes"`
= `126/8568`
= `1/68`
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