Question

A problem in mathematics is given to three students whose chances of solving it are $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$​
$\left( i \right)$ what is the probability that the problem is solved?
$\left( {ii} \right)$ What is the probability that exactly one of them will solve it?

Answer 1

First we have to define what the terms we need to solve in the problem are.
Since there are three students are solving the math’s problem into some probabilities like students who are all chances of solving the given set of problems are $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$ ​and we need to find the overall probability and also the probability of the exactly one of them will need to solve it. Hence, we use the concept of the probability which is the number of favorable events that divides the total number of events.

Complete step-by-step solution:
Since from the given question there are three students taken as A, B, and C. and also from the given question, the probability of the student A can solves the given question is $\dfrac{1}{3}$ which is $p(A) = \dfrac{1}{3}$
Similarly, further finding all other students in the same way, for student B $p(B) = \dfrac{1}{4}$and students C is $p(C) = \dfrac{1}{5}$ (probability that C can solve the given problem)
Hence $\left( i \right)$ what is the probability that the problem is solved is we need to subtract all known values into the minus one and hence, $\Rightarrow [1 - p(A)][1 - p(B)][1 - p(C)]$ now substitute the know values we get
$\Rightarrow [1 - p(A)][1 - p(B)][1 - p(C)] = (1 - \dfrac{1}{3})(1 - \dfrac{1}{4})(1 - \dfrac{1}{5})$ ( cross multiply and solve)
$\Rightarrow \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} = \dfrac{2}{5}$ (which is the probability that the problem cannot be solved). So we do one minus the probability of not solving the problem.
Hence for the problem can be solved is $1 - \dfrac{2}{5} = \dfrac{3}{5}$ (the problem can be solved as in the probability)
$\left( {ii} \right)$ What is the probability that exactly one of them will solve it?
Since exactly one needs to answer, so the probability is $p({A} \cap {B’} \cap C’ or A’ \cap {B} \cap {C’} or {A’} \cap B’ \cap {C})$ (power to the dash means the one minus the original values so that he will cannot answer)
We used the concept of possible counting so in first C will answer and A and B will not answer, hence similarly
Thus $p({A} \cap {B’} \cap {C’} or A’ \cap {B} \cap {C’} or {A’} \cap {B’} \cap {C}) = \dfrac{1}{3}\times\dfrac{3}{4} \times \dfrac{4}{5}+\dfrac{2}{3}\times\dfrac{1}{4} \times \dfrac{4}{5}+\dfrac{2}{3}\times\dfrac{3}{4} \times \dfrac{1}{5}= \dfrac{{12}}{{60}} + \dfrac{8}{{60}} + \dfrac{6}{{60}} = \dfrac{{13}}{{30}}$(thus the probability of one exactly solves the math problem)

Note: Since the $probability{\text{ }} = {\text{ }}\dfrac{{number{\text{ }}of{\text{ }}favorable outcomes}}{{total{\text{ }}number of outcomes}}{\text{ }}$ to that we find the favorable events in the above and divides will get the required answer; also $p({A^1} \cap {B^1} \cap CorA \cap {B^1} \cap {C^1}or{A^1} \cap B \cap {C^1})$ is the formula for exactly one (need to answer); using the concept of probability, and counting principle.