When dealing with probabilities in problems involving conditional events, remember to use the law of total probability. In this case, break down the probability of heads into two cases: selecting a double-headed coin and selecting a fair coin, and then sum the probabilities.
The correct answer is: (C): 10
We are given a bag containing \( 2n + 1 \) coins. Out of these, \( n \) coins have heads on both sides (double-headed), and the remaining \( n + 1 \) coins are fair. One coin is selected at random and tossed. The probability that the toss results in heads is \( \frac{31}{42} \). We are tasked with finding the value of \( n \).
Step 1: Probability of selecting a coin
The probability of selecting a double-headed coin is \[ \frac{n}{2n + 1} \] and the probability of selecting a fair coin is \[ \frac{n + 1}{2n + 1}. \]
Step 2: Probability of getting heads
For a double-headed coin, the probability of getting heads is 1. For a fair coin, the probability of getting heads is \[ \frac{1}{2}. \]
Step 3: Total probability of getting heads
By the law of total probability: \[ P(\text{Heads}) = \frac{n}{2n + 1} \times 1 + \frac{n + 1}{2n + 1} \times \frac{1}{2} \] Simplifying: \[ P(\text{Heads}) = \frac{n}{2n + 1} + \frac{n + 1}{2(2n + 1)} \]
Step 4: Equating to the given probability
Since \( P(\text{Heads}) = \frac{31}{42} \), we equate: \[ \frac{n}{2n + 1} + \frac{n + 1}{2(2n + 1)} = \frac{31}{42} \] Combining like terms: \[ \frac{2n + (n + 1)}{2(2n + 1)} = \frac{31}{42} \] This simplifies to: \[ \frac{3n + 1}{2(2n + 1)} = \frac{31}{42} \]
Step 5: Solve for \( n \)
Cross-multiplying: \[ 42(3n + 1) = 62(2n + 1) \] Expanding: \[ 126n + 42 = 124n + 62 \] Simplifying: \[ 2n = 20 \quad \Rightarrow \quad n = 10 \]
Conclusion:
The value of \( n \) is 10, so the correct answer is (C): 10.