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.
If a random variable \( x \) has the probability distribution 
then \( P(3<x \leq 6) \) is equal to
Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant, and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage, and radish to be 25%, 35%, and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2