Question

A die, whose faces are marked $1,2,3$ in red and $4,5,6$ in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find if A and B are independent events.

Answer 1

First in a die, there are six faces which are one, two, three, four, five, and six in the numbers form.
Die is in the shape of a square. From the given that red is marked as one, two, three in the die, and green is marked as four, five, six in the given die.
Dependent events mean taking the product of two probability sets is equal to the intersection of the probability of the two sets.

Complete step-by-step solution:
After the die is tossed, From the given event A is the obtained even numbers which are $\\{ 2,4,6\\}$ from the given six numbers in the die.
And the event B is the obtained red which is $\\{ 1,2,3\\}$.
Now we are going the find the probability of the events, using the probability formula that $\text{Probability} = \dfrac{\text{Favorable}}{\text{Total}}$ (favorable events is the possible outcome and total is the overall given values)
Thus, we get for the probability of the event A is $P(A) = \dfrac{3}{6}$ (three even numbers are the favorable events a total we have six faces)
Further solving we get, $P(A) = \dfrac{3}{6} \Rightarrow \dfrac{1}{2}$
Probability of the event B is $P(B) = \dfrac{3}{6}$ (three red marked numbers is the favorable events a total we have six faces)
Further solving we get, $P(B) = \dfrac{3}{6} \Rightarrow \dfrac{1}{2}$
Now intersecting the events A and B we get, $A \cap B = \\{ 2,4,6\\} \cap \\{ 1,2,3\\}$ (from the given events)
Thus, we get the intersection of A and B, $A \cap B = \\{ 2,4,6\\} \cap \\{ 1,2,3\\} \Rightarrow 2$(common elements from both sets)
Now taking the probability for the intersection of A and B, we get $P(A \cap B) = \dfrac{1}{6}$ (one is intersection numbers is the favorable events a total we have six faces)
Finally, we need to check if A and B are independent events,
If A and B are dependent then, $P(A \cap B) = P(A) \times P(B)$
Substitute the values that we found above, we get, $P(A \cap B) = P(A) \times P(B) \Rightarrow \dfrac{1}{6} = \dfrac{3}{6} \times \dfrac{3}{6}$
Further solving we get, $P(A \cap B) = P(A) \times P(B) \Rightarrow \dfrac{1}{6} \ne \dfrac{1}{4}$ and hence A and B are independent as $P(A \cap B) \ne P(A) \times P(B)$

Note: If suppose $P(A \cap B) = P(A) \times P(B)$ then the events A and B are dependent.
Since the total outcome is six because if a die is tossed there are total outcomes in that.
In the probability of event A, $P(A) \Rightarrow \dfrac{1}{2}$ if we divide this and multiplied by a hundred, we get the percentage of the event A, $P(A) = \dfrac{1}{2} \times 100 \Rightarrow 50\%$