JEE MAIN - Mathematics (2002 - No. 26)
A problem in mathematics is given to three students $$A,B,C$$ and their respective probability of solving the problem is $${1 \over 2},{1 \over 3}$$ and $${1 \over 4}.$$ Probability that the problem is solved is :
$${3 \over 4}$$
$${1 \over 2}$$
$${2 \over 3}$$
$${1 \over 3}$$
Explanation
Given $$P\left( A \right) = {1 \over 2}$$, $$P\left( B \right) = {1 \over 3}$$, $$P\left( C \right) = {1 \over 4}$$
So, $$P\left( {\overline A } \right) = {1 \over 2}$$ (Probablity that the problem can't be solve by A)
$$P\left( {\overline B } \right) = {2 \over 3}$$ (Probablity that the problem can't be solve by B)
and $$P\left( {\overline C } \right) = {3 \over 4}$$ (Probablity that the problem can't be solve by C)
Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C
$$P\left( {A \cup B \cup C} \right)$$ = 1 - $$P\left( {\overline A } \right)P\left( {\overline B } \right)P\left( {\overline C } \right)$$
$$ = 1 - {1 \over 2} \times {2 \over 3} \times {3 \over 4}$$
$$=$$ $${3 \over 4}$$
So, $$P\left( {\overline A } \right) = {1 \over 2}$$ (Probablity that the problem can't be solve by A)
$$P\left( {\overline B } \right) = {2 \over 3}$$ (Probablity that the problem can't be solve by B)
and $$P\left( {\overline C } \right) = {3 \over 4}$$ (Probablity that the problem can't be solve by C)
Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C
$$P\left( {A \cup B \cup C} \right)$$ = 1 - $$P\left( {\overline A } \right)P\left( {\overline B } \right)P\left( {\overline C } \right)$$
$$ = 1 - {1 \over 2} \times {2 \over 3} \times {3 \over 4}$$
$$=$$ $${3 \over 4}$$
Comments (0)
