Question

A problem in mathematics is given to 3 students whose chances of solving individually are \[\dfrac{1}{2}\], \[\dfrac{1}{3}\] and \[\dfrac{1}{4}\]. The probability that the problem will be solved at least by one, is ?

A) \[\dfrac{1}{4}\]
B) \[\dfrac{1}{24}\]
C) \[\dfrac{23}{24}\]
D) \[\dfrac{3}{4}\]

Answer 1

Hint: To solve this question we will use basic formulas of probability which is given as probability of not happening of event is given by, \[P\left( \overline{A} \right)=\left[ 1-P\left( A \right) \right]\], where P(A) is the probability of happening of A and \[P\left( \overline{A} \right)\] is the probability of not happening of A.
Complete step-by-step answer:
It is given in the question that a problem in mathematics is given to 3 students whose chances of solving individually are \[\dfrac{1}{2}\], \[\dfrac{1}{3}\] and \[\dfrac{1}{4}\].
We need to find the probability that the question is solved by at least one of them.
We will first make certain assumptions to solve the question.
Let A be the first student with probability P(A) in solving the problem.
Similarly, B be the second student with probability P(B) in solving the question.
And lastly C be the third student with probability P(C) in solving the question.
substituting the above given values in the probability term we get,
P(A) = \[\dfrac{1}{2}\]​, P(B) = \[\dfrac{1}{3}\] and P(C) = ​\[\dfrac{1}{4}\]
Let us assume E as the event that the problem will be solved at least by 1.
Then this probability is equal to 1 minus the product of the probabilities of not happening of A, B and C.
Probability of not happening of an event A is given by \[P\left( \overline{A} \right)\]
Therefore, the probability of event E is given by \[=~1-[P\left( \overline{A} \right)P\left( \overline{B} \right)P\left( \overline{C} \right)]\]


\[\Rightarrow \] The probability of event E \[=~1-[P\left( \overline{A} \right)P\left( \overline{B} \right)P\left( \overline{C} \right)]\]
Substituting the value of \[P\left( \overline{A} \right)=\left[ 1-P\left( A \right) \right]\] and similarly of others we get,

\[\Rightarrow \] The probability of event E \[=~~1-[\left[ 1-P\left( A \right) \right]\left[ 1-P\left( B \right) \right]\left[ 1-P\left( C \right) \right]]\]

\[\Rightarrow \] The probability of event E \[=~~1-\left[ 1-\dfrac{1}{2} \right]\left[ 1-\dfrac{1}{3} \right]\left[ 1-\dfrac{1}{4} \right]\]
\[\Rightarrow \] The probability of event E = \[\left[ 1-\dfrac{1}{4} \right]\]
\[\Rightarrow \] The probability of event E \[=\dfrac{3}{4}\].
Therefore, the answer to the above question is \[\dfrac{3}{4}\] i.e. option (D).

Note: The possibility of error in these types of questions can be at the point of confusion of whether to add or multiply the required probabilities. We always add the probabilities when they are independent to each other and multiply them when they are dependent.