Question

The probability of solving a problem by three students A , B and C are $\dfrac{1}{2},\dfrac{3}{4}{\text{ and }}\dfrac{1}{4}$
Respectively. The probability that the problem will be solved is
${\text{A}}{\text{. }}\dfrac{3}{{32}} \\\ {\text{B}}{\text{. }}\dfrac{3}{{16}} \\\ {\text{C}}{\text{. }}\dfrac{{29}}{{32}} \\\ {\text{D}}{\text{. None of the above}} \\\$

Answer 1

Hint: -You have probability of solving problems by students A, B and C is given you have find probability that problem will be solved will be equal to 1- probability of problem can’t be solved.

Complete step-by-step answer:
Given
$P\left( A \right) = \dfrac{1}{2},P\left( B \right) = \dfrac{3}{4},P\left( C \right) = \dfrac{1}{4}$
$\Rightarrow P\left( {\vec A} \right) = 1 - P\left( A \right) \\\ \therefore P\left( {\vec A} \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\\$
$\Rightarrow P\left( {\vec B} \right) = 1 - P\left( B \right) \\\ \Rightarrow P\left( {\vec B} \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4} \\\$
$\Rightarrow P\left( {\vec C} \right) = 1 - P\left( C \right)$
$\therefore P\left( {\vec C} \right) = 1 - \dfrac{1}{4} = \dfrac{1}{4}$
The probability that problem will be solved = 1- (probability that problem can’t be solved).
$\therefore$Required probability = 1- $P\left( {\vec A \cap \vec B \cap \vec C} \right)$
$\therefore 1 - \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{{29}}{{32}}$
Hence option C is the correct option.
Note: -Whenever you get this type of question the key concept of solving is if you have to find probability of success you can write it 1- probability of failure. And you have knowledge of $P\left( {\vec A \cap \vec B \cap \vec C} \right) = P\left( {\vec A} \right) \times P\left( {\vec B} \right) \times P\left( {\vec C} \right)$.