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.
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.
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:
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is