Question

Let A and B are two independent events. The probability that both A and B occur together is $\dfrac{1}{6}$and the probability that either of them occurs is $\dfrac{1}{3}$. The probability of occurrence of A is
$1) $$0$or $1$
$2) $$\dfrac{1}{2}$or $\dfrac{1}{3}$
$3) $$\dfrac{1}{2}$or $\dfrac{1}{4}$
$4) $$\dfrac{1}{3}$or $\dfrac{1}{4}$

Answer 1

Hint: First, we need to know the concept of probability.
Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
Like the probability of the A and B occur together is given as $\dfrac{1}{6}$which means the favorable event is $1$and the total outcome is $6$
 Formula used:
$P = \dfrac{F}{T}$where P is the probability, F is the possible favorable events and T is the total outcomes from the given.
For independent events with two values $P(A \cap B)$can be expressed in the form of $P(A \cap B) = P(A) \times P(B)$
$P({A^{-1}})$is the probability of event A not occurring and thus we get $P({A^{-1}}) = 1 - P(A)$is the probability of event A occurring where $1$is the total probability and any functions will not exceed $1$.

Complete step by step answer:
From the given that probability of both A and B occur together is $\dfrac{1}{6}$and this can be expressed in mathematically as $P(A \cap B) = \dfrac{1}{6}$(A and B compared)
Also, given that, the probability that either of them occurs is $\dfrac{1}{3}$. If A occurs then B will not, and similarly if B occurs then A will not. This can be expressed as \[P({A^{-1}} \cap {B^1}) = \dfrac{1}{3}\]where $P({A^{-1}})$is the probability of event A not occurring and $P({B^1})$is the probability of event B not occurring and the intersection given the probability as in the given.
Since A and B are independent events, $P(A \cap B) = P(A) \times P(B)$ (for the intersection)
Applying this in the above equation we get, $P(A \cap B) = P(A) \times P(B) \Rightarrow \dfrac{1}{6}$
Similarly, A and B are independent events, thus \[P({A^{-1}} \cap {B^1}) = P({A^{-1}}) \times P({B^1}) \Rightarrow \dfrac{1}{3}\]
This equation can be rewritten by using the $P({A^{-1}}) = 1 - P(A)$is the probability of event A occurring where $1$is the total probability.
Thus, we get \[P({A^{-1}}) \times P({B^1}) = [1 - P(A)] \times [1 - P(B)] \Rightarrow \dfrac{1}{3}\]
Now the use of multiplication operation, we get \[[1 - P(A)] \times [1 - P(B)] = 1 + P(A)P(B) - P(A) - P(B) \Rightarrow \dfrac{1}{3}\] (where $(1 - a)(1 - b) = 1 + ab - a - b$))
Substituting the value of the intersection of A and B we get, \[1 + P(A)P(B) - P(A) - P(B) = \dfrac{1}{3} \Rightarrow 1 + \dfrac{1}{6} - P(A) - P(B) = \dfrac{1}{3}\] where $P(A \cap B) = P(A) \times P(B) \Rightarrow \dfrac{1}{6}$
Further solving the equation, we get, \[1 + \dfrac{1}{6} - P(A) - P(B) = \dfrac{1}{3} \Rightarrow - P(A) - P(B) = \dfrac{1}{3} - \dfrac{7}{6}\]
\[ \Rightarrow - P(A) - P(B) = \dfrac{1}{3} - \dfrac{7}{6} \Rightarrow P(A) + P(B) = \dfrac{5}{6}\]
Again, apply this value in $P(A) \times P(B) \Rightarrow \dfrac{1}{6}$we get; $P(A)[\dfrac{5}{6} - P(A)] \Rightarrow \dfrac{1}{6}$ where \[P(B) = \dfrac{5}{6} - P(A)\]
Hence solving this we get, $P(A)[\dfrac{5}{6} - P(A)] = \dfrac{1}{6} \Rightarrow 6P{(A)^2} - 5P(A) + 1 = 0$
By the quadratic formula we get, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$where $a = 6,b = - 5,c = 1$
Hence, we get; $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{5 \pm \sqrt {{{( - 5)}^2} - 4(6)} }}{{2(6)}} \Rightarrow \dfrac{{5 \pm 1}}{{12}}$
The quadratic values are $\dfrac{6}{{12}},\dfrac{4}{{12}} \Rightarrow \dfrac{1}{2},\dfrac{1}{3}$

So, the correct answer is “Option 2”.

Note: The quadratic formulas are applied to the degree two equations.
We are also able to solve this problem by converting into $P(A) = x,P(B) = y$so that there will be no confusion while solving the degree two-quadratic equations.
The total probability value will not exceed the one and $P({A^{-1}}) = 1 - P(A)$is the complement of the given event.