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 the problem is solved is:
\( \frac{5}{8} \)
The probability that none of them solves the problem is:
\[ P(\text{None}) = (1 - P_1) (1 - P_2) (1 - P_3) \]
Substituting given values:
\[ P(\text{None}) = \left( 1 - \frac{2}{5} \right) \left( 1 - \frac{1}{4} \right) \left( 1 - \frac{1}{6} \right) \]
\[ = \frac{3}{5} \times \frac{3}{4} \times \frac{5}{6} \]
\[ = \frac{3 \times 3 \times 5}{5 \times 4 \times 6} = \frac{45}{120} = \frac{3}{8} \]
The probability that at least one of them solves the problem is:
\[ P(\text{At least one}) = 1 - P(\text{None}) \]
\[ = 1 - \frac{3}{8} = \frac{5}{8} \]
Thus, the correct answer is option (c).
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 either only Kalyani or Kashvi or Sara solves it 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 : ?
