Kashvi, Kalyani, and Sara of class 12 are given a problem in Accounts whose respective probabilities of solving are \( \frac{2}{5}, \frac{1}{4}, \) and \( \frac{1}{6} \). They were asked to solve it independently.
The probability that either only Kalyani or Kashvi or Sara solves it is:
\( \frac{9}{20} \)
To find the probability that exactly one of them solves the problem, we consider three cases:
- Only Kashvi solves it.
- Only Kalyani solves it.
- Only Sara solves it.
Each case follows the formula:
\[ P(\text{Only one solves}) = P(\text{Person solves}) \times P(\text{Others fail}) \]
### Case 1: Only Kashvi solves
\[ P(\text{Only Kashvi}) = P(Kashvi) \times (1 - P(Kalyani)) \times (1 - P(Sara)) \]
\[ = \frac{2}{5} \times \left(1 - \frac{1}{4}\right) \times \left(1 - \frac{1}{6}\right) \]
\[ = \frac{2}{5} \times \frac{3}{4} \times \frac{5}{6} \]
\[ = \frac{30}{120} = \frac{1}{4} \]
### Case 2: Only Kalyani solves
\[ P(\text{Only Kalyani}) = P(Kalyani) \times (1 - P(Kashvi)) \times (1 - P(Sara)) \]
\[ = \frac{1}{4} \times \left(1 - \frac{2}{5}\right) \times \left(1 - \frac{1}{6}\right) \]
\[ = \frac{1}{4} \times \frac{3}{5} \times \frac{5}{6} \]
\[ = \frac{15}{120} = \frac{1}{8} \]
### Case 3: Only Sara solves
\[ P(\text{Only Sara}) = P(Sara) \times (1 - P(Kashvi)) \times (1 - P(Kalyani)) \]
\[ = \frac{1}{6} \times \left(1 - \frac{2}{5}\right) \times \left(1 - \frac{1}{4}\right) \]
\[ = \frac{1}{6} \times \frac{3}{5} \times \frac{3}{4} \]
\[ = \frac{9}{120} = \frac{3}{40} \]
### Total Probability:
\[ P(\text{Exactly one solves}) = \frac{1}{4} + \frac{1}{8} + \frac{3}{40} \]
\[ = \frac{10}{40} + \frac{5}{40} + \frac{3}{40} = \frac{18}{40} = \frac{9}{20} \]
Thus, the correct answer is option (a).
Similar Questions
Which of the following is the probability of \( x \) successes in a binomial distribution with number of trials \( n \) and probability of success as \( \theta \) ( \( 0 < \theta < 1 \) ) in each trial?
Let \( X \) be a random variable whose probability distribution is given by the table:
X 1 3 5 7 P(X) \( \frac{1}{3} \) \( \frac{1}{6} \) \( \frac{1}{6} \) \( \frac{1}{3} \) Then variance of \( X \) is:
For a Binomial distribution \( B(n, p) \), \( \frac{E(x)}{V(x)} \) is equal to:
The probability that the problem is solved is:
Important Questions from Probability
Rakesh is 17th from the right and Ankit is 15th from the left in a line of students. If they interchange their places, the position of Ankit becomes 19th from the left. How many students are there in the line?
What comes in place of the question mark (?) in the series given below?
B2D, C3F, E5J, G7N, ?, M13Z
If 1st January, 2001 was a Monday, what was the day on 26th January, 2003?
From the given options, at what angle are the hands of a clock inclined at 10 minutes to 2 (Smaller angle)?
In the given analogy, choose the number which will replace the question mark (?).
WSH : 5 : : KMJ : ?
