Question

Find the probability that either A or B can solve a problem when chosen randomly if A can solve $75\%$ of the problems in Mathematics where B can solve $70\%$ respectively.
(a) $\dfrac{{37}}{{40}}$
(b) $\dfrac{{19}}{{40}}$
(c) $\dfrac{{39}}{{40}}$
(d) Cannot be determined

Answer 1

Hint : The given problem revolves around the concepts of probability. So, we will use the definition of probability for each outcome or for the given outcome particularly. Using the formula for the probability i.e., if the probability of any event O is denoted by ‘P(O)’ (in this case particularly) where the formula exists that $P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( s \right)}}$, where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event. Then, using the formula $p\left( {A \cup B} \right) = \left[ {p\left( A \right)} \right]\left[ {p\left( {{{\bar B}}} \right)} \right] + \left[ {p\left( B \right)} \right]\left[ {p\left( {{{\bar A}}} \right)} \right] + \left[ {p\left( A \right)} \right]\left[ {p\left( {{B}} \right)} \right]$ (where, ‘$\cup$’ denotes the probability of union of two possibilities in the respective event) to find probability in both the case, the required outcome is obtained.

Complete step-by-step answer :
Since, we have given the two different events or say, class namely ‘$A$’ and ‘$B$’ which can solve problems of mathematics particularly which are probable of getting $75\%$ and $70\%$ respectively;
$\therefore$We know that,
Probability of getting the outcome(s) for any possibilities i.e.
$Probability = \dfrac{{number{{ }}of{{ }}outcomes}}{{total{{ }}outcomes{{ }}in{{ }}the{{ }}event}} = \dfrac{{n\left( s \right)}}{{n\left( O \right)}}$ … ($1$)
Where,
In this case particularly, total outcomes in the event are, $n\left( O \right) = 100$ (since, outcomes given in percentage, $\%$)
As a result, by the given condition using the equation ($1$), we can solve for getting the required value
$\therefore$Probability of solving problems for ‘$A$’ is,
$Probability,{{ }}p\left( A \right) = \dfrac{{75}}{{100}}$
Simplifying the equation mathematically i.e. multiplying and dividing by $25$, we get
$p\left( A \right) = \dfrac{3}{4}$ … (i)
Similarly,
Probability of ‘$A$’ for not solving the problems in Mathematics (expressed as ${{\bar A}}$) is,
$Probability,{{ }}p\left( {{{\bar A}}} \right) = 1 - p\left( A \right)$
Simplifying the equation mathematically i.e. substituting the value of $p\left( A \right)$, we get
$p\left( {{{\bar A}}} \right) = 1 - \dfrac{3}{4} = \dfrac{{4 - 3}}{4}$
$p\left( {{{\bar A}}} \right) = \dfrac{1}{4}$ … (ii)
Also,
For the probability of solving problems for ‘$B$’ is,
$Probability,{{ }}p\left( B \right) = \dfrac{{70}}{{100}}$
Simplifying the equation mathematically i.e. multiplying and dividing by $10$, we get
$p\left( B \right) = \dfrac{7}{{10}}$ … (iii)
Similarly,
Probability of ‘$B$’ for not solving the problems in Mathematics (expressed as ${{\bar B}}$) is,
$Probability,{{ }}p\left( {{{\bar B}}} \right) = 1 - p\left( B \right)$
Simplifying the equation mathematically i.e. substituting the value of $p\left( B \right)$, we get
$p\left( {{{\bar B}}} \right) = 1 - \dfrac{7}{{10}} = \dfrac{{10 - 7}}{{10}}$
$p\left( {{{\bar B}}} \right) = \dfrac{3}{{10}}$ … (iv)
Now,
Hence, from (i), (ii), (iii) and (iv) respectively
It seems that,
Probability of solving the problems by A or B class is found to be from $p\left( {A \cup B} \right) = \left[ {p\left( A \right)} \right]\left[ {p\left( {{{\bar B}}} \right)} \right] + \left[ {p\left( B \right)} \right]\left[ {p\left( {{{\bar A}}} \right)} \right] + \left[ {p\left( A \right)} \right]\left[ {p\left( {{B}} \right)} \right]$
As a result, substituting the values in the above equation, we get
$p\left( {A \cup B} \right) = \left( {\dfrac{3}{4}} \right)\left( {\dfrac{3}{{10}}} \right) + \left( {\dfrac{7}{{10}}} \right)\left( {\dfrac{1}{4}} \right) + \left( {\dfrac{3}{4}} \right)\left( {\dfrac{7}{{10}}} \right)$
Solving the equation predominantly, we get
$p\left( {A \cup B} \right) = \dfrac{9}{{40}} + \dfrac{7}{{40}} + \dfrac{{21}}{{40}}$
As a result, adding the whole equation (as the denominator is same), we get
$p\left( {A \cup B} \right) = \dfrac{{37}}{{40}}$
$\therefore \Rightarrow$The option (a) is correct.
So, the correct answer is “Option a”.

Note : One should know the basic formula of probability while solving a question. We should also know the concepts of independent events and that their probabilities are not affected by each other’s occurrences. We should take care of the calculations so as to be sure of our final answer. Also, we should know that probabilities to find the possibilities in two cases (like in such situation) is always $p\left( {A \cup B} \right) = \left[ {p\left( A \right)} \right]\left[ {p\left( {{{\bar B}}} \right)} \right] + \left[ {p\left( B \right)} \right]\left[ {p\left( {{{\bar A}}} \right)} \right] + \left[ {p\left( A \right)} \right]\left[ {p\left( {{B}} \right)} \right]$ respectively.