There are 52 runners in a race. How many ways can the runners finish first, second, and third?
1 answer:
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Answer:
(a)
(b)
(c)
Step-by-step explanation:
Given
i.e. between 4 and 9
Solving (a): Probability that each of the 5 digit are different
Since there is no restriction;
The total possible selection is as follows:
(i.e. any of the 4 start digits)
(i.e. any of the 10 digits 0 - 9)
(i.e. any of the 10 digits 0 - 9)
(i.e. any of the 10 digits 0 - 9)
(i.e. any of the 10 digits 0 - 9)
So, the total is:
For selection that all digits are different, the selection is:
(i.e. any of the 4 start digits)
(i.e. any of the remaining 9)
(i.e. any of the remaining 8)
(i.e. any of the remaining 7)
(i.e. any of the remaining 6)
So:
So, the probability is:
Solving (b): At least 1 repeated digit
The probability calculated in (a) is the all digits are different i.e. P(None)
So, using laws of complement
We have:
So, we have:
Solving (c): An expression to model the probability.
<em>Using (a) as a point of reference, we have;</em>
Where
---- for selection of 5 i.e. n = 5
This can be rewritten as:
4 can be expressed as: 5 - 1
So, we have:
Substitute n for 5
This can be rewritten as:
So, the expression is:
<em>Where n represents the digit number</em>