Given the probability that A can solve a problem is $\dfrac{2}{3}$ and the probability that B can solve the same problem is $\dfrac{3}{5}$. Find the probability that none of two will be able to solve the problem.

Hint: Here we solve the problem by finding the individual probability of events for not solving the problem because here the events have solved the problem individually. If suppose $p(x)$is the probability of solving some x work then 1-p(x) will be the probability for not solving work.

Complete step-by-step answer:

Now here let us consider that A be an event that solves problem A and also B be an event that solves problem B.

According to the data we can say that
 Probability that the event A can solve problem A is $\dfrac{2}{3}$ $ \Rightarrow P(A) = \dfrac{2}{3}$
Probability that the event B can solve the problem B is$\dfrac{3}{5}$ $ \Rightarrow P(B) = \dfrac{3}{5}$

Here we have to find the probability that none of two will be able to solve the problem.

So here we have to find the value of $P(\bar A.\bar B)$.
\[ \Rightarrow P(\bar A.\bar B) = P(\bar A)P(\bar B)\]

Since we know that the event A and B has solved the problem individually so we should also find their individual probability of not solving the problem.

\[ \Rightarrow P(\bar A.\bar B) = P(\bar A)P(\bar B)\]
$ \Rightarrow P(\bar A.\bar B) = (1 - P(A))(1 - P(B))$
$ \Rightarrow P(\bar A.\bar B) = \left( {1 - \dfrac{2}{3}} \right)\left( {1 - \dfrac{3}{5}} \right)$
$ \Rightarrow P(\bar A\bar B) = \dfrac{1}{3} \times \dfrac{2}{5}$
$ \Rightarrow P(\bar A\bar B) = \dfrac{2}{{15}}$

Therefore the probability that none of the two events A and B will be able to solve the problem = $\dfrac{2}{{15}}$.

Note: In this problem there are two events A and First we have to observe that event A and event B are working independently, which means the probability of event A solving (or not solving) the problem is entirely independent of the probability of event B solving (or not solving) the problem. Based on this we have to find the values and use them according to it.