Question

Suppose that we roll a red and a black die. Let a = "the black die shows a 2 or a 5", b = "the sum of the two dice is at least 7". Are events a and b independent?

Answer 1

Answer: No, A and B are not independent events.

∵ it does not satisfy the rule of probability for independent events i.e.

P(A∩B)=P(A).P(B)

Explanation:

Let A be the event that  the black dice shows 2 or 5

Let B be the event that the sum of two dice is atleast 7

Sample space of A={$(R_1,B_2)(R_2,B_2)(R_3,B_2)(R_4,B_2)$($(R_5,B_2),(R_6,B_2),(R_1,B_5),(R_2,B_5),(R_3,B_5),(R_4,B_5),(R_5,B_5),(R_6,B_5)$}

Sample space of B= { $(R_1,B_6),(R_2,B_5),(R_2,B_6),(R_3,B_4),(R_3,B_5),(R_3,B_6),$,$(R_4,B_3),(R_4,B_4),(R_4,B_5),(R_4,B_6),(R_5,B_2),(R_5,B_3),(R_5,B_4),(R_5,B_5)$, $(R_5,B_6),(R_6,B_2),(R_6,B_3),(R_6,B_4),(R_6,B_5),(R_6,B_5)$}

P(A)= $\frac{\text{No.of favourable outcomes}}{\text{Total no.of observation}}$

⇒P(A)=$\frac{12}{36}$

⇒P(A)=$\frac{1}{3}$

Similarly,

P(B)=$\frac{20}{36}$

⇒ P(B) =$\frac{5}{9}$

Now for Sample Space of (A∩B)= {$(R_5,B_2)(R_6,B_2)(R_2,B_5)(R_3,B_5)(R_4,B_5)(R_5,B_5)(R_6,B_5)$}

So, P(A∩B)= $\frac{7}{36}$

Now we apply the formula,

P(A).P(B)=P(A∩B)

$\frac{1}{3}$×$\frac{5}{9}$ ≠ $\frac{7}{36}$

$\frac{5}{27}$ ≠ $\frac{7}{36}$

∴ The events A and B are not independent events.