A two-digit number is written down at random.Find the probability that the number will be
A)Smaller than 20
B)Even
C)Multiple of 5
number of two-digit numbers:
10, 11, ... , 99, ----> 90 of them (1 - 9 are missing)
smaller than 20:
10,11, ... , 19 ----> 10 of them
prob( < 20) = 10/90 = 1/9
use the same type of analysis for the two other parts, let me know what you get.
10/90
=1/9
Nice
It will to self study
To find the probability of an event occurring, we need to divide the number of favorable outcomes by the total number of possible outcomes. Let's consider each part separately:
A) Smaller than 20:
In this case, we need to determine the number of two-digit numbers smaller than 20. Those numbers are:
10, 11, 12, 13, 14, 15, 16, 17, 18, and 19.
Total possible outcomes: There are 90 two-digit numbers (10 to 99).
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 10 / 90
Probability = 1 / 9
Therefore, the probability that a two-digit number will be smaller than 20 is 1/9.
B) Even:
To determine the number of even two-digit numbers, we need to consider the digits in the ones place. The even digits are: 0, 2, 4, 6, and 8. Since the tens place can be any digit (0 to 9), we have a total of 5 even digits available for the ones place.
Total possible outcomes: There are 90 two-digit numbers (10 to 99).
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 5 / 90
Probability = 1 / 18
Therefore, the probability that a two-digit number will be even is 1/18.
C) Multiple of 5:
To determine the number of two-digit multiples of 5, we need to consider the digits in the ones place. The multiples of 5 are: 5, 10, 15, 20,..., 95.
Total possible outcomes: There are 90 two-digit numbers (10 to 99).
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 18 / 90
Probability = 1 / 5
Therefore, the probability that a two-digit number will be a multiple of 5 is 1/5.