You forget the last two digits of your
cell phone password.
a. What is the probability that you randomly choose
the correct digits?
b. Suppose you remember that both digits are even. How
does this change the probability that you choose the
correct digits?
To answer these questions, we need to consider the total number of possible combinations and the number of favorable outcomes.
a. If you forget the last two digits of your cell phone password, assuming it is a four-digit password (e.g., XXXX), we can calculate the probability of randomly choosing the correct digits.
Step 1: Determine the total number of possible combinations.
In this case, the last two digits can each take on values from 0 to 9, so there are 10 options for each digit. Therefore, there are a total of 10 * 10 = 100 possible combinations.
Step 2: Determine the number of favorable outcomes.
Since you forgot the last two digits, there is only one correct combination. So, the number of favorable outcomes is 1.
Step 3: Calculate the probability.
The probability (P) is calculated as the number of favorable outcomes divided by the total number of possible combinations:
P = Number of Favorable Outcomes / Total Number of Possible Combinations
P = 1/100 = 0.01
Therefore, the probability of randomly choosing the correct digits is 0.01 or 1%.
b. Now, let's consider the scenario where you remember that both digits are even (i.e., the last two digits can only be 0, 2, 4, 6, or 8).
Step 1: Determine the total number of possible combinations.
Since each digit can be one of the five even numbers, there are 5 options for each digit. Therefore, there are a total of 5 * 5 = 25 possible combinations.
Step 2: Determine the number of favorable outcomes.
In this case, there are two even digits remaining (e.g., if the password was 4X6X, it could be 40, 42, 44, 46, or 48), so the number of favorable outcomes is 5 * 5 = 25.
Step 3: Calculate the probability.
P = Number of Favorable Outcomes / Total Number of Possible Combinations
P = 25/25 = 1
Therefore, the probability of randomly choosing the correct digits when you know that both digits are even is 1 or 100%.
I googled "number of possible passcodes" and it told be
10,000 which is 10^4, so there are no restrictions
your prob of the event you are after = 10^2/10^4 = 1/100
Assuming you count 0 as "even" , then you only have 5 choices at each of the last places
prob = 5^2/10^2 = 25/10000= 1/3600
otherwise, if 0 is not counted as even
prob = 16/10000 = 1/5625
you decide about the 0