Answer :
Let's solve this problem step-by-step.
### Step 1: Understanding the scenario
The question is asking for the probability that a group of 5 random digits chosen from a given string of digits will contain at least 3 odd digits.
### Step 2: Count the occurrences of odd and even digits
We'll use the given string of digits:
```
46370551705348049126802127520267201882419780838154
```
Count the occurrences of odd and even digits:
- Odd digits: 1, 3, 5, 7, 9
- Even digits: 0, 2, 4, 6, 8
According to the counted values, we have:
- Odd count: 21
- Even count: 29
### Step 3: Calculate probabilities of picking odd and even digits
Calculate the probability of picking an odd digit (P(odd)) and an even digit (P(even)):
- Total digits: 50 (sum of odd and even digits)
Then the probabilities will be:
- P(odd) = odd count / total digits = 21/50 ≈ 0.42
- P(even) = even count / total digits = 29/50 ≈ 0.58
### Step 4: Using binomial distribution for calculation
We need the probability that out of 5 random digits, at least 3 will be odd.
Using the binomial distribution formula, where [tex]\( n \)[/tex] is the number of trials (5), [tex]\( k \)[/tex] is the number of successes (at least 3), and [tex]\( p \)[/tex] is the probability of success (0.42 for odd digits):
The binomial probability [tex]\( P(X = k) \)[/tex] is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot (p)^k \cdot (1-p)^{n-k} \][/tex]
We sum these probabilities for [tex]\( k = 3, 4, 5 \)[/tex]:
[tex]\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \][/tex]
### Step 5: Result
After performing these calculations, the probability that a group of 5 random digits will contain at least 3 odd digits is approximately:
[tex]\[ \approx 0.3525 \][/tex]
### Step 6: Conclusion
Therefore, the correct probability from the given multiple choices:
A. [tex]\(\frac{4}{5}\)[/tex] (0.80)
B. [tex]\(\frac{3}{5}\)[/tex] (0.60)
C. [tex]\(\frac{9}{5}\)[/tex] (1.80)
D. [tex]\(\frac{1}{1}\)[/tex] (1.00)
None of these answer choices accurately reflect the probability based on our calculation of 0.3525, which indicates there might be a mistake or typo in the provided answer choices. The correct choice is not listed among the options provided.
### Step 1: Understanding the scenario
The question is asking for the probability that a group of 5 random digits chosen from a given string of digits will contain at least 3 odd digits.
### Step 2: Count the occurrences of odd and even digits
We'll use the given string of digits:
```
46370551705348049126802127520267201882419780838154
```
Count the occurrences of odd and even digits:
- Odd digits: 1, 3, 5, 7, 9
- Even digits: 0, 2, 4, 6, 8
According to the counted values, we have:
- Odd count: 21
- Even count: 29
### Step 3: Calculate probabilities of picking odd and even digits
Calculate the probability of picking an odd digit (P(odd)) and an even digit (P(even)):
- Total digits: 50 (sum of odd and even digits)
Then the probabilities will be:
- P(odd) = odd count / total digits = 21/50 ≈ 0.42
- P(even) = even count / total digits = 29/50 ≈ 0.58
### Step 4: Using binomial distribution for calculation
We need the probability that out of 5 random digits, at least 3 will be odd.
Using the binomial distribution formula, where [tex]\( n \)[/tex] is the number of trials (5), [tex]\( k \)[/tex] is the number of successes (at least 3), and [tex]\( p \)[/tex] is the probability of success (0.42 for odd digits):
The binomial probability [tex]\( P(X = k) \)[/tex] is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot (p)^k \cdot (1-p)^{n-k} \][/tex]
We sum these probabilities for [tex]\( k = 3, 4, 5 \)[/tex]:
[tex]\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \][/tex]
### Step 5: Result
After performing these calculations, the probability that a group of 5 random digits will contain at least 3 odd digits is approximately:
[tex]\[ \approx 0.3525 \][/tex]
### Step 6: Conclusion
Therefore, the correct probability from the given multiple choices:
A. [tex]\(\frac{4}{5}\)[/tex] (0.80)
B. [tex]\(\frac{3}{5}\)[/tex] (0.60)
C. [tex]\(\frac{9}{5}\)[/tex] (1.80)
D. [tex]\(\frac{1}{1}\)[/tex] (1.00)
None of these answer choices accurately reflect the probability based on our calculation of 0.3525, which indicates there might be a mistake or typo in the provided answer choices. The correct choice is not listed among the options provided.
