Question

The odds against a certain event are 5 to 2, and the odds in favour of another event independent of the former are 6 to 5: find the chance that one at least of the events will happen.

Answer 1

Hint- This question is of probability chapter where the main concept for this question is to find the probability of at least one condition, in that case we simply find the probability when both conditions do not occur and just we subtract it from 1 to get the at least one event must occur.

Complete step-by-step solution -
Let the first event whose odds against are 5 to 2 ${E_1}$ and the second event whose odd in favour are 6 to 5 be ${E_2}$
As mentioned in the question the odds against the first event is 5 to 2
Therefore the probability of the first event is given as the number of favorable events upon the number of total events.
Therefore of probability of happening the first event is
$
  P\left( {{E_1}} \right) = \dfrac{{favourable\,\,outcomes}}{{total\,number\,of\,outcomes}} \\
  P\left( {{E_1}} \right) = \dfrac{2}{{5 + 2}} = \dfrac{2}{7} \\
$
Since we have to find the probability of not happening the first event, which we can calculate by subtracting the probability of happening the first event from 1
Probability of not happening the first event is
$
  P\left( {{{\overline E }_1}} \right) = 1 - P\left( {{E_1}} \right) \\
  P\left( {{{\overline E }_1}} \right) = 1 - \dfrac{2}{7} \\
  P\left( {{{\overline E }_1}} \right) = \dfrac{5}{7} \\
$
For the second case it is given that the odds in favor of happening the event are 6 to 5
Therefore of probability of happening the second event is
$
  P\left( {{E_2}} \right) = \dfrac{{favourable\,\,outcomes}}{{total\,number\,of\,outcomes}} \\
  P\left( {{E_2}} \right) = \dfrac{6}{{6 + 5}} = \dfrac{6}{{11}} \\
$
Similarly, we have to find the probability of not happening the second event, which we can calculate by subtracting the probability of happening the second event from 1
Probability of not happening the second event is
$
  P\left( {{{\overline E }_2}} \right) = 1 - P\left( {{E_2}} \right) \\
  P\left( {{{\overline E }_2}} \right) = 1 - \dfrac{6}{{11}} \\
  P\left( {{{\overline E }_2}} \right) = \dfrac{5}{{11}} \\
$
As we have calculated the probability of not happening the two events separately, now we will calculate the probability of happening at least one event
$
  P\left( E \right) = 1 - \left( {P\left( {\overline {{E_1}} } \right) \times P\left( {\overline {{E_1}} } \right)} \right) \\
  P\left( E \right) = 1 - \left( {\dfrac{5}{7} \times \dfrac{5}{{11}}} \right) \\
  P\left( E \right) = 1 - \dfrac{{25}}{{77}} \\
  P\left( E \right) = \dfrac{{52}}{{77}} \\
$
Hence, the required probability is $P\left( E \right) = \dfrac{{52}}{{77}}$.

Note- In order to solve these types of problems, learn the basic definition of probability. The first step is to declare the events and then calculate the probability of the events and proceed step by step to solve the problem. The above problem can also be solved with the help of Venn diagram. The steps are defining the universal set and the events are defined as subsets of the universal set and then solve the problem using laws of sets.