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
A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throws, and a number less than 4 in the third throw?
Two dice are thrown simultaneously. If \( X \) denotes the number of fours, then the expectation of \( X \) will be:
If the random variable \( X \) has the following distribution:
X 0 1 2 otherwise P(X) k 2k 3k 0 Match List-I with List-II:
List-I List-II (A) k (I) \(\frac{5}{6}\) (B) P(X < 2) (II) \(\frac{4}{3}\) (C) E(X) (III) \(\frac{1}{2}\) (D) P(1 ≤ X ≤ 2) (IV) \(\frac{1}{6}\) Choose the correct answer from the options given below:
Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form, where c is some constant:
\[ P(X = x) = \begin{cases} 0.1, & \text{if } x = 0 \\ cx, & \text{if } x = 1 \text{ or } x = 2 \\ c(5 - x), & \text{if } x = 3 \text{ or } x = 4 \\ 0, & \text{otherwise} \end{cases} \]
Match List-I with List-II:
List-I List-II (A) c (I) 0.75 (B) P(X ≤ 2) (II) 0.3 (C) P(X = 2) (III) 0.55 (D) P(X ≥ 2) (IV) 0.15 Choose the correct answer from the options given below:
For the differential equation \( (x \log_e x) dy = (\log_e x - y) dx \):
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.
(D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where A is an arbitrary constant.
Choose the correct answer from the options given below:
