1 point
A and B predicts the outcomes of a cricket match and their chances of predicting the runs scored by a specific batsman correctly are
3
6
6
3
​
and
2
3
3
2
​
respectively independent of each other. If the probability of them predicting the same wrong score is
5
215
215
5
​
. Given that they predicted the same score, find the probability that their answer is correct.

To solve the problem, we need to determine the probability that A and B predicted the correct score given that they both predicted the same score.

### Step 1: Define the Probabilities

Let's define the probabilities for A and B predicting correctly and incorrectly.

- \( P(A \text{ correct}) = \frac{3}{6} = \frac{1}{2} \)
- \( P(A \text{ incorrect}) = 1 - \frac{1}{2} = \frac{1}{2} \)
- \( P(B \text{ correct}) = \frac{2}{3} \)
- \( P(B \text{ incorrect}) = 1 - \frac{2}{3} = \frac{1}{3} \)

### Step 2: Calculate the Joint Probability of Both Predicting the Same Wrong Score

We are given:
\[ P(A \text{ and } B \text{ same wrong score}) = \frac{5}{215} \]

### Step 3: Calculate the Probability of Both Predicting the Same Score

We need the total probability that both predict the same score (whether correct or incorrect).

- Probability that both predict the same correct score:
  \[
  P(A \text{ correct}) \times P(B \text{ correct}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}
  \]

- Probability that both predict the same wrong score is already given:
  \[
  P(\text{same wrong score}) = \frac{5}{215}
  \]

Total probability of both predicting the same score:
\[
P(\text{same score}) = P(\text{same correct score}) + P(\text{same wrong score}) = \frac{1}{3} + \frac{5}{215}
\]

To add these fractions, we need a common denominator:
\[
\frac{1}{3} = \frac{215}{645}
\]
\[
\frac{5}{215} = \frac{15}{645}
\]

So,
\[
P(\text{same score}) = \frac{215}{645} + \frac{15}{645} = \frac{230}{645}
\]

### Step 4: Calculate the Conditional Probability

We need to find \( P(\text{correct} \mid \text{same score}) \):
\[
P(\text{correct} \mid \text{same score}) = \frac{P(\text{same correct score})}{P(\text{same score})}
\]

Substitute the known values:
\[
P(\text{correct} \mid \text{same score}) = \frac{\frac{1}{3}}{\frac{230}{645}} = \frac{1}{3} \times \frac{645}{230} = \frac{645}{690} = \frac{215}{230}
\]

Simplify the fraction:
\[
\frac{215}{230} = \frac{215 \div 5}{230 \div 5} = \frac{43}{46}
\]

Thus, the probability that their answer is correct given they predicted the same score is:
\[
\boxed{\frac{43}{46}}
\]

Would you like more details or have any questions?

### Related Questions:

1. What is the probability that A predicts the score correctly but B predicts it incorrectly?
2. If A and B predict independently, what is the combined probability that at least one of them predicts correctly?
3. How would the probability change if A's accuracy increased to \(\frac{4}{6}\)?
4. What is the expected number of correct predictions by A in a series of 10 matches?
5. How does conditional probability help in real-world decision-making?
6. If B's accuracy dropped to \(\frac{1}{3}\), how would it affect the overall probability calculations?
7. Can the probability of them predicting the same score (either correct or incorrect) ever exceed 1?
8. How would you use Bayes' theorem to solve this problem?

### Tip:

When dealing with conditional probabilities, always identify the total probability of the given condition to simplify the calculations.

Explore the probability of correct predictions in a cricket match scenario where two predictors, A and B, independently predict the same score. Understand how conditional probability is applied to determine the likelihood of their answer being correct given their prediction.

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