To determine the probability that a group of 5 random digits contains at least 2 even digits in the given sequence, we need to follow these steps:
1. Identify the total number of 5-digit groups that can be formed:
- Given the string of 50 digits: "46370551705348049126892127529267291882413780838154", the total number of 5-digit groups we can form is the number of possible starting positions for these groups.
- In total, there are [tex]\(50 - 5 + 1 = 46\)[/tex] such groups (since the first group starts at the first digit, the second group starts at the second digit, and so on, until the last group starts at the 46th digit).
2. Count the number of groups with at least 2 even digits:
- Even digits are [tex]\( \{0, 2, 4, 6, 8\} \)[/tex].
- We check each of the 46 groups of 5 digits to determine how many of them contain at least 2 even digits.
The counting process reveals that out of the 46 groups, there are 40 groups that contain at least 2 even digits.
3. Calculate the probability:
- Probability is defined as the number of favorable outcomes divided by the total number of possible outcomes.
- The number of favorable outcomes (groups with at least 2 even digits) is 40.
- The total number of possible groups is 46.
So the probability [tex]\( P \)[/tex] is:
[tex]\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{40}{46} \][/tex]
Converting this fraction into its decimal form:
[tex]\[ P = \frac{40}{46} \approx 0.869565 \][/tex]
4. Simplify and match with provided answers:
- Converting [tex]\( \frac{40}{46} \)[/tex] to a simpler fraction: The greatest common divisor of 40 and 46 is 2, so:
[tex]\[ \frac{40 \div 2}{46 \div 2} = \frac{20}{23} \][/tex]
- This fraction is not exactly represented among the choices, so the next step is to compare its decimal form to the closest fraction among the given options.
Among the provided options:
A. [tex]\( \frac{7}{10} = 0.7 \)[/tex]
B. [tex]\( \frac{9}{10} = 0.9 \)[/tex]
C. [tex]\( \frac{3}{5} = 0.6 \)[/tex]
D. [tex]\( \frac{4}{5} = 0.8 \)[/tex]
The closest approximation to [tex]\( \frac{20}{23} \approx 0.869565 \)[/tex] is option B, [tex]\( \frac{9}{10} \)[/tex].
Therefore, the correct answer corresponding to the probability that a group of 5 random digits contains at least 2 even digits is:
B. [tex]\( \frac{9}{10} \)[/tex]
