The probabilities of A, B and C solving a problem independently are \frac{1}{2},\, \frac{1}{3}\: and\; \frac{1}{4}  respectively. If all the three try to solve the problem independently, find the probability that the problem is solved.

 

 

 

 
 
 
 
 

Answers (1)

P\left ( A \right )= \frac{1}{2}            P\left ( B \right )= \frac{1}{3}           P\left ( C \right )= \frac{1}{4}
P\left ( \bar{A} \right )= \frac{1}{2}           P\left ( \bar{B} \right )= \frac{2}{3}           P\left ( \bar{c} \right )= \frac{3}{4}
P\left ( \bar{A}\cap\bar{B}\cap \bar{C} \right )= P\left ( \bar{A} \right )\cdot P\left ( \bar{B} \right )P\left ( \bar{C} \right )\Rightarrow P\left ( \bar{S} \right )---(1)
P\left ( S \right )= 1-P\left ( \bar{S} \right )
P\left ( S \right )=  probability of solving a problem
P\left ( \bar{S} \right )= probability of not solving a problem
P\left ( \bar{S} \right )= P\left ( \bar{A}\cap \bar{B}\cap \bar{C} \right )= \frac{1}{\not{2}}\times \frac{\not{2}}{\not{3}}\times \frac{\not{3}}{4}\: \: \left [ from(1) \right ]
                                                = \frac{1}{4}
P\left ( S \right )= 1-\frac{1}{4}\Rightarrow \frac{4-1}{4}= \frac{3}{4} 
 

Preparation Products

Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions