Question

A can solve $90\%$ of problems given in a book and B can solve $70\%$.What is the probability that at least one of them will solve the problem, selected at random from the book?

Answer 1

We are given the chances of solving a problem of both A and B. Using this we can find the probabilities of the event A solving the problem and B solving the problem. Then we need to find the probability that A will not be able to do the problem and the same for B. We need to find their product and subtract it from one to get the required probability.

Complete step-by-step answer:
We are given that A can solve $90\%$ of problems given in a book and B can solve $70\%$.
So now we can consider the happening of two events E and F
E is the event that A solves the problem selected at random
F is the event that B solves the problem selected at random
The probability of an event A is given by the formula $\dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Here since we are given in percentage.
We get the probability of the occurrence of event E to be $P\left( E \right) = \dfrac{{90}}{{100}}$ as A can solve $90\%$ of problems
Now, we get the probability of the occurrence of event F to be $P\left( F \right) = \dfrac{{70}}{{100}}$ as B can solve $70\%$ of problems
Now let’s find the probability that A cannot solve the problem, which is given by subtracting $P\left( E \right)$ from $1$
$P\left( {\overline E } \right) = 1 - P\left( E \right) = 1 - \dfrac{{90}}{{100}} = \dfrac{{10}}{{100}}$
Now let’s find the probability that B cannot solve the problem, which is given by subtracting $P\left( F \right)$ from $1$
$P\left( {\overline F } \right) = 1 - P\left( F \right) = 1 - \dfrac{{70}}{{100}} = \dfrac{{30}}{{100}}$
The product of these two probabilities gives the probability of which both A and B cannot solve the problem.
$P\left( {\overline E } \right).P\left( {\overline F } \right) = \dfrac{{10}}{{100}}.\dfrac{{30}}{{100}} = \dfrac{{300}}{{10000}} = \dfrac{3}{{100}}$
Now we need the probability that either one of them will solve the problem, which is nothing but $\overline {P\left( {\overline E } \right).P\left( {\overline F } \right)}$
Subtracting $P\left( {\overline E } \right).P\left( {\overline F } \right)$ from $1$ we get the required probability$\overline {P\left( {\overline E } \right).P\left( {\overline F } \right)} = 1 - P\left( {\overline E } \right).P\left( {\overline F } \right) = 1 - \dfrac{3}{{100}} = \dfrac{{97}}{{100}}$

Note: The same problem can be done using another formula of probability. From the question , we need to find the probability that at least one of them will solve the problem which means either one needs to solve that is nothing but $P(E \cup F)$.This is given by the formula $P(E \cup F) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)$ where $P\left( {E \cap F} \right)$ is the product of the probabilities.