Question

If A and B are two independent events such that \[P(\bar A \cap B) = \dfrac{2}{{15}}\] and \[P(A \cap \bar B) = \dfrac{1}{6}\] , then find P(A) and P(B).

Answer 1

Hint:
The probability of an event tells us how likely two possible events is to occur. The probability of an event is the number of favourable outcomes divided by the total number of outcomes.

Step-wise-solution:
Given, A and B are two independent events \[P(\bar A \cap B) = \dfrac{2}{{15}}\] and \[P(A \cap \bar B) = \dfrac{1}{6}\] .
To find P(A) and P(B).
We know, \[P(A \cap \bar B) = \dfrac{1}{6}\], and we also know that \[P(A \cap \bar B) = P(A)P(\bar B)\]
Therefore, \[P(A)P(\bar B) = \dfrac{1}{6}\]
\[ \Rightarrow P(A)[1 - P(B)] = \dfrac{1}{6}\]
$ \
  { \Rightarrow P(A) = \dfrac{1}{{6[1 - P(B)]}}} \\ $ , which can be written as equation (i).
Similarly, since it is given \[P(\bar A \cap B) = \dfrac{2}{{15}}\]
Therefore, we can write \[ \Rightarrow P(B)[1 - P(A)] = \dfrac{2}{{15}}\]
Putting P(A) from equation (i), \[P(B)[1 - \dfrac{1}{{6[1 - P(B)]}}] = \dfrac{2}{{15}}\]
\[ \Rightarrow 30P{(B)^2} + 29P(B) - 4 = 0\]
Multiplying the above equation with (-1), we get,
\[ \Rightarrow 30P{(B)^2} - 29P(B) + 4 = 0\]
Let P(B) = x for simplicity
Then, we find the above equation to be like the one mentioned below, i.e.
\[ \Rightarrow 30{x^2} - 29x + 4 = 0\]
 \[ \Rightarrow (6x - 1)(5x - 4) = 0\]
Now, there are two cases.
Case-I:
Either \[6x - 1 = 0\]
\[ \Rightarrow x = \dfrac{1}{6}\], \[P(B) = \dfrac{1}{6}\]
and \[P(A) = \dfrac{1}{5}\]
Case-II:
\[5x - 4 = 0\]
\[x = \dfrac{5}{6}\], that is, \[P(B) = \dfrac{5}{6}\]
and \[P(A) = \dfrac{4}{5}\]
Hence, the answers are \[P(A) = \dfrac{1}{5}\] , \[P(B) = \dfrac{1}{6}\]
or, \[P(B) = \dfrac{5}{6}\], \[P(A) = \dfrac{4}{5}\]



Note:
Thus, by solving the question, you must have got an idea that you need the knowledge of probability to solve both questions. Also, it is possible that there is more than one value to this question because we are provided with data in the same way. With this, remember the formulas of probability so that you can approach these questions easily.