Question
Question: A two-digit number from \[10\] to \[99\], inclusive, is chosen at random. What is the probability th...
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) 51
B) 92
C) 9010
D) 1918
E) 9119
Solution
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:
Probability=Total number of outcomesNumber 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.
⇒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:
⇒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.
⇒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:
⇒Probability=9018
By dividing into the RHS side of the expression Probability=9018 , we get:
⇒Probability=51
∴ 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⩽P(A)⩽1 .
Also, you should remember that the number of outcomes will always be less than the total number of outcomes.