
Three numbers are chosen at random from 1 to 20. The probability that they are consecutive is
A. $\dfrac{1}{190}$
B. $\dfrac{1}{120}$
C. $\dfrac{3}{190}$
D. $\dfrac{5}{190}$
Answer
494.1k+ views
Hint: We first explain the concept of empirical probability and how the events are considered. We take the given events and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the shooting event.
Complete step by step answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We find the number of outcomes for both events. We take the conditional event A as the chosen three numbers to be consecutive and the universal event as U as choosing 3 numbers and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$ respectively.
We take the empirical probability of the given problem as $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$.
We are choosing 3 consecutive numbers from 1 to 20. If we take them as sets, the first number of the set can be 1 to 18. So, there are 18 possible choices for consecutive three numbers. $n\left( A \right)=18$.
Now we chose three numbers out of 20 numbers and $n\left( U \right)={}^{20}{{C}_{3}}=\dfrac{20\times 19\times 18}{6}=1140$.
The empirical probability of the shooting event is $P\left( A \right)=\dfrac{18}{1140}=\dfrac{3}{190}$.
So, the correct answer is “Option C”.
Note: We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.
Complete step by step answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We find the number of outcomes for both events. We take the conditional event A as the chosen three numbers to be consecutive and the universal event as U as choosing 3 numbers and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$ respectively.
We take the empirical probability of the given problem as $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$.
We are choosing 3 consecutive numbers from 1 to 20. If we take them as sets, the first number of the set can be 1 to 18. So, there are 18 possible choices for consecutive three numbers. $n\left( A \right)=18$.
Now we chose three numbers out of 20 numbers and $n\left( U \right)={}^{20}{{C}_{3}}=\dfrac{20\times 19\times 18}{6}=1140$.
The empirical probability of the shooting event is $P\left( A \right)=\dfrac{18}{1140}=\dfrac{3}{190}$.
So, the correct answer is “Option C”.
Note: We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.
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