Question

A problem in mathematics is given to 4 students whose chances of solving individually are $\dfrac{1}{2}$, $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$. The probability that the problem will be solved at least by one student is?
(a) $\dfrac{2}{3}$
(b) $\dfrac{3}{5}$
(c) $\dfrac{4}{5}$
(d) $\dfrac{3}{4}$

Answer 1

Hint: First, we will assign the name of the 4 students and also assign the events of them solving the questions. Then, we will find the complementary events of those probabilities. By using the condition of intersection of complements, we will find the required probability.

Complete step by step answer:
Here, there is a mathematics question which needs to be solved by 4 students, let us say Student A, Student B, Student C and Student D and the probabilities of the students solving the question is $\dfrac{1}{2}$, $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$ respectively.
First, let us assign the events to the students solving the mathematics question and find their complementary events which is given by, $P\left( \overline{E} \right)=1-P\left( E \right)$, similarly find out for the other events.
Let E be the event that student A solves the question, therefore the probability of A solving the question, $P\left( E \right)=\dfrac{1}{2}$ and their complementary is $P\left( \overline{E} \right)=\dfrac{1}{2}$.
Let F be the event that student B solves the question, therefore the probability of B solving the question, $P\left( F \right)=\dfrac{1}{3}$ and their complementary is $P\left( \overline{F} \right)=\dfrac{2}{3}$.
Let G be the event that student C solves the question, therefore the probability of C solving the question, $P\left( G \right)=\dfrac{1}{4}$ and their complementary is $P\left( \overline{G} \right)=\dfrac{3}{4}$.
Let H be the event that student D solves the question, therefore the probability of D solving the question, $P\left( H \right)=\dfrac{1}{5}$ and their complementary is $P\left( \overline{H} \right)=\dfrac{4}{5}$.
To find probability that the problem will be solved at least by one student,
We know, by intersection of complements
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$
Now, here the 1 represents the total probability of success and failure combined, and $P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$ represents that out of all those 4 students none of them have solved that question, so when we solve the above expression, we will get at least one student solves the question, $P\left( E\cup F\cup G\cup H \right)$. We know, that events $E$, $F$, $G$ and $H$ are independent events, therefore, $\overline{E}$, $\overline{F}$, $\overline{G}$ and $\overline{H}$ are also independent events respectively.
Since, we know the intersection of the independent sets is equal to multiplication of their respective probabilities, we get
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E} \right)\cdot P\left( \overline{F} \right)\cdot P\left( \overline{G} \right)\cdot P\left( \overline{H} \right)$
\[\begin{align}
  & =1-\left( \dfrac{1}{2} \right)\cdot \left( \dfrac{2}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
 & =1-\left( 1 \right)\cdot \left( \dfrac{1}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
 & =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
 & =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{5} \right) \\
 & =1-\dfrac{1}{5} \\
 & =\dfrac{1\times 5}{1\times 5}-\dfrac{1}{5} \\
 & =\dfrac{5-1}{5} \\
 & =\dfrac{4}{5} \\
\end{align}\]

Hence, the probability that the problem will be solved at least by one student is $\dfrac{4}{5}$.

Note: Probability is numerical mathematics where we can tell how often an event can happen. The sum of the probabilities of all possible outcomes is 1. A probability of an event is a likelihood of that event occurring.