A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $ 1,2,3,4,5,6,7,8 $ and these are equally likely outcomes. What is the probability that it will point at
A) 8?
B) An odd number?
C) A number greater than 2?
D) A number less than 9?
Answer
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Hint: First we will find the total number of chances a spinning arrow consists, which is $ 8 $ and is clear from the question. Then one by one we solve all parts. For each part we separately calculate the favorable possible outcomes and then, use the formula of probability to get the answer.
General formula of probability is $ P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $ n\left( S \right)= $ number of total possible outcomes
Complete step-by-step answer:
We have given that a game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $ 1,2,3,4,5,6,7,8 $
So, total number of possible outcomes $ n\left( S \right)=8 $
Now, let’s solve part A
We have to find the probability that the arrow will point at $ 8 $
As $ 8 $ is the only number on the wheel, so the number of favorable outcomes will be $ 1 $
So, the probability of getting $ 8 $ will be
$ P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $
$ P\left( 8 \right)=\dfrac{1}{8} $
Let’s solve part B
We have to find the probability that the arrow will point at an odd number.
The odd numbers from $ 1,2,3,4,5,6,7,8 $ are \[1,3,5,7\]
So, the number of favorable outcomes will be $ n\left( E \right)=4 $ .
Probability of getting an odd number will be
$ \begin{align}
& P\left( \text{odd} \right)=\dfrac{4}{8} \\
& P\left( \text{odd} \right)=\dfrac{1}{2} \\
\end{align} $
Let’s solve part C
We have to find the probability that the arrow will point at a number greater than $ 2 $ .
The numbers greater than $ 2 $ are $ 3,4,5,6,7,8 $
So, the number of favorable outcomes will be $ n\left( E \right)=6 $
So, the probability of getting a number greater than $ 2 $ will be
$ \begin{align}
& P\left( \text{greater than }2 \right)=\dfrac{6}{8} \\
& P\left( \text{greater than }2 \right)=\dfrac{3}{4} \\
\end{align} $
Let’s solve part D
We have to find the probability that the arrow will point at a number less than $ 9 $ .
The numbers less than $ 9 $ are $ 1,2,3,4,5,6,7,8 $
Total of favorable outcomes will be $ n\left( E \right)=8 $
So, the probability of getting a number less than $ 9 $ will be
$ \begin{align}
& P\left( \text{less than 9} \right)=\dfrac{8}{8} \\
& P\left( \text{less than 9} \right)=1 \\
\end{align} $
Note: One should know about odd numbers to solve the question correctly are. Odd numbers are not divisible by $ 2 $ . Also, while solving part C, we should not include $ 2 $ in our possible outcomes as we have to find the probability of getting a number greater than $ 2 $ . Probability value $ 1 $ indicates $ 100% $ chances. The value of probability cannot exceed one. Probability value higher than one means probability greater than $ 100% $ and it is not possible.
General formula of probability is $ P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $ n\left( S \right)= $ number of total possible outcomes
Complete step-by-step answer:
We have given that a game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $ 1,2,3,4,5,6,7,8 $
So, total number of possible outcomes $ n\left( S \right)=8 $
Now, let’s solve part A
We have to find the probability that the arrow will point at $ 8 $
As $ 8 $ is the only number on the wheel, so the number of favorable outcomes will be $ 1 $
So, the probability of getting $ 8 $ will be
$ P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $
$ P\left( 8 \right)=\dfrac{1}{8} $
Let’s solve part B
We have to find the probability that the arrow will point at an odd number.
The odd numbers from $ 1,2,3,4,5,6,7,8 $ are \[1,3,5,7\]
So, the number of favorable outcomes will be $ n\left( E \right)=4 $ .
Probability of getting an odd number will be
$ \begin{align}
& P\left( \text{odd} \right)=\dfrac{4}{8} \\
& P\left( \text{odd} \right)=\dfrac{1}{2} \\
\end{align} $
Let’s solve part C
We have to find the probability that the arrow will point at a number greater than $ 2 $ .
The numbers greater than $ 2 $ are $ 3,4,5,6,7,8 $
So, the number of favorable outcomes will be $ n\left( E \right)=6 $
So, the probability of getting a number greater than $ 2 $ will be
$ \begin{align}
& P\left( \text{greater than }2 \right)=\dfrac{6}{8} \\
& P\left( \text{greater than }2 \right)=\dfrac{3}{4} \\
\end{align} $
Let’s solve part D
We have to find the probability that the arrow will point at a number less than $ 9 $ .
The numbers less than $ 9 $ are $ 1,2,3,4,5,6,7,8 $
Total of favorable outcomes will be $ n\left( E \right)=8 $
So, the probability of getting a number less than $ 9 $ will be
$ \begin{align}
& P\left( \text{less than 9} \right)=\dfrac{8}{8} \\
& P\left( \text{less than 9} \right)=1 \\
\end{align} $
Note: One should know about odd numbers to solve the question correctly are. Odd numbers are not divisible by $ 2 $ . Also, while solving part C, we should not include $ 2 $ in our possible outcomes as we have to find the probability of getting a number greater than $ 2 $ . Probability value $ 1 $ indicates $ 100% $ chances. The value of probability cannot exceed one. Probability value higher than one means probability greater than $ 100% $ and it is not possible.
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