
A two-digit number from \[10\] to \[99\], inclusive, is chosen at random. What is the probability that this number is divisible by \[5\] ?
A) \[\dfrac{1}{5}\]
B) \[\dfrac{2}{9}\]
C) \[\dfrac{{10}}{{90}}\]
D) \[\dfrac{{18}}{{19}}\]
E) \[\dfrac{{19}}{{91}}\]
Answer
573k+ views
Hint: Here we will use the formula for finding the probability which states that the probability for occurring any event will be equals to Number of favorable outcomes divided by the total number of favorable/possible outcomes:
\[{\text{Probability}} = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Complete step-by-step solution:
Step 1: First of all, we will calculate the total number that exists \[99\]. If we start from \[10\] and end with inclusive \[99\] , the total number is \[90\].
\[ \Rightarrow \]The total number of outcomes is equal to \[90\].
Step 2: Now, we will find out how many numbers are there between \[10\] \[99\] which are completely divisible by \[5\] because we need to find the probability of that number which is divisible by \[5\].
As we know, a number is divisible by
\[5\] only when it ends with either
\[0\] or \[5\].
So, the numbers divisible by \[5\] from \[10\] to \[99\] are as below:
\[ \Rightarrow 10\], \[15\], \[20\], \[25\], \[30\], \[35\], \[40\],\[45\], \[50\],\[55\], \[60\],\[65\], \[70\],\[75\],\[80\],\[85\],\[90\],\[95\] .
There is a total \[18\] number which is divisible by \[5\].
\[ \Rightarrow \]The number of favorable outcomes is equal to \[90\].
Step 3: Now, we will use the probability formula for calculating the answer. By substituting the values of the number of outcomes and the total number of outcomes, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{{{\text{18}}}}{{{\text{90}}}}\]
By dividing into the RHS side of the expression \[{\text{Probability}} = \dfrac{{{\text{18}}}}{{{\text{90}}}}\] , we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{{\text{1}}}{5}\]
\[\therefore \] Option A is correct.
Note: Students need to remember some basic points about the probability that the range of the probability will always lie between \[0 \leqslant {\text{P(A)}} \leqslant {\text{1}}\] .
Also, you should remember that the number of outcomes will always be less than the total number of outcomes.
\[{\text{Probability}} = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Complete step-by-step solution:
Step 1: First of all, we will calculate the total number that exists \[99\]. If we start from \[10\] and end with inclusive \[99\] , the total number is \[90\].
\[ \Rightarrow \]The total number of outcomes is equal to \[90\].
Step 2: Now, we will find out how many numbers are there between \[10\] \[99\] which are completely divisible by \[5\] because we need to find the probability of that number which is divisible by \[5\].
As we know, a number is divisible by
\[5\] only when it ends with either
\[0\] or \[5\].
So, the numbers divisible by \[5\] from \[10\] to \[99\] are as below:
\[ \Rightarrow 10\], \[15\], \[20\], \[25\], \[30\], \[35\], \[40\],\[45\], \[50\],\[55\], \[60\],\[65\], \[70\],\[75\],\[80\],\[85\],\[90\],\[95\] .
There is a total \[18\] number which is divisible by \[5\].
\[ \Rightarrow \]The number of favorable outcomes is equal to \[90\].
Step 3: Now, we will use the probability formula for calculating the answer. By substituting the values of the number of outcomes and the total number of outcomes, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{{{\text{18}}}}{{{\text{90}}}}\]
By dividing into the RHS side of the expression \[{\text{Probability}} = \dfrac{{{\text{18}}}}{{{\text{90}}}}\] , we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{{\text{1}}}{5}\]
\[\therefore \] Option A is correct.
Note: Students need to remember some basic points about the probability that the range of the probability will always lie between \[0 \leqslant {\text{P(A)}} \leqslant {\text{1}}\] .
Also, you should remember that the number of outcomes will always be less than the total number of outcomes.
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