Question

Roll a die and consider the following two events={2, 3, 6}, event ={1, 5, 6}. Are the events E and F independent?

Answer 1

Answer:

Option 1 - No, the probability of F occurring is higher if it is known that E has occurred.

Step-by-step explanation:

Given : Roll a die and consider the following two events={2, 3, 6}, event ={1, 5, 6}.

To find : Are the events E and F independent?

Solution :

E={2, 3, 6}, F={1, 5, 6}

They are not independent event as 6 is common in both the events.

Now, verifying by applying independent property,

Probability of event E is $P(E)=\frac{3}{6}=\frac{1}{2}$

Probability of event F is $P(F)=\frac{3}{6}=\frac{1}{2}$

Probability of E and F is $P(E\cap F)=\frac{1}{6}$

Probability of F occuring when it is known that e has occured.

i.e. $P(F/E)=\frac{P(E\cap F)}{P(E)}$

$P(F/E)=\frac{\frac{1}{6}}{\frac{1}{2}}$

$P(F/E)=\frac{1}{3}$

$P(F)>P(F/E)$

i.e. The probability of F occurring is higher if it is known that E has occurred.

Therefore, The correct option is 1.

Events E and F are not independent, because  the probability of F occurring is higher if it is known that E has occurred.

Answer 2

Answer:  The correct option is

(B) No, because there is at least one roll which leads to both E and F occurring.

Step-by-step explanation:  We are given to roll a die and consider the following two events :

A = {2, 3, 6}  and  B = {1, 5, 6}.

We are to check whether the events A and B are independent.

We have

A ∩ B = {2, 3, 6} ∩ {1, 5, 6} = {6}.

So, n(A) = 3,  n(B) = 3  and  n(A ∩ B) = 1.

Since there are 6 elements in the sample space, we get

S = {1, 2, 3, 4, 5, 6}   ⇒   n(S) = 6.

So, the probabilities of the events A, B and A ∩ B are calculated as follows :

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2},\\\\\\P(B)=\dfrac{n(B)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2},\\\\\\P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{1}{6}.$

Therefore,

$P(A\cap B)=\dfrac{1}{6}\neq P(A)\times P(B)=\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}$

Thus, the events A and B are NOT independent because at least one roll which leads to both E and F occurring  and that is 6, common to both A and B.

Option (B) is CORRECT.