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ser-zykov [4K]
3 years ago
14

A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons

ist of all odd digits if the same digit is not used more than once in the code?
Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
5 0
The answer is D: <span>4,920 out of 5,040</span>
Eduardwww [97]3 years ago
4 0
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

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Sladkaya [172]

Answer:

a) 0.1423

b) 0.2977

c) 0.56

Step-by-step explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that p = 0.13

14 drivers are stopped

This means that n = 14

(a) None will have an invalid license.

This is P(X = 0)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{14,0}.(0.13)^{0}.(0.87)^{14} = 0.1423

(b) Exactly one will have an invalid license.

This is P(X = 1)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{14,1}.(0.13)^{1}.(0.87)^{13} = 0.2977

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

P(X < 2) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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