To determine the probabilities for the two parts of your question, let's analyze them one at a time.
Question 1
a. Each digit in the password can be any number from 0 to 9, giving us a total of 10 possible choices for each digit. If you forgot the last two digits, the total number of combinations for the two digits is:
\[ 10 \times 10 = 100 \]
If you are randomly choosing two digits, only one of those combinations is correct. Therefore, the probability of randomly selecting the correct two digits is:
\[ \frac{1}{100} = 0.01 \]
To express this as a percent, we multiply by 100:
\[ 0.01 \times 100 = 1% \]
So, the probability that you randomly choose the correct digits is 1%.
Question 2
The probability as a percent is:
1%.
Question 3
b. If you remember that both digits are even, you are reducing your options. The even digits available are 0, 2, 4, 6, and 8, making a total of 5 possible choices for each digit. Thus, the total number of combinations for the two even digits is:
\[ 5 \times 5 = 25 \]
Since there is still only one correct combination, the new probability of randomly selecting the correct even digits is:
\[ \frac{1}{25} \]
To express this as a percent, we multiply by 100:
\[ \frac{1}{25} \times 100 = 4% \]
So, it increases the probability that your choice is correct to 1 over 25, or 4% because each digit could be 0, 2, 4, 6, or 8.
Thus, the correct response from the provided options is:
"It increases the probability that your choice is correct to 1 over 25, or 4%, because each digit could be 0, 2, 4, 6, or 8."