Question

5. a) Consider the case of tossing two dices. Let

: A be the event of getting an odd total

: B the event of getting on the first dice, and

: C the event of getting a total of seven on two dices.

Then

(i) Find (P A ∩C), (P A ∩ )B and (P B∩ C).

Check whether A,B and C are independent or not. Give reasons for your answers.

Answer 1

Solution:

$n(S)=36$

$A=\{(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),(4,3)\\ (4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5) \}$

So, $n(A)=18$

Since B is incomplete in question, we assume B be the event of getting 1 on the first dice.

$B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\}$

So, $n(B)=6$

$C=\{(1,6),(6,1),(2,5)(5,2),(3,4),(4,3)\}$

So, $n(C)=6$

$A\cap B=\{(1,2),(1,4),(1,6)\}$

So, $n(A\cap B)=3$

$B\cap C=\{(1,6)\}$

So, $n(B\cap C)=1$

$A\cap C=\{(1,6),(6,1),(2,5)(5,2),(3,4),(4,3)\}$

So, $n(A\cap C)=6$

(i) Now, $P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{3}{36}=\dfrac1{12}$

$P(B\cap C)=\dfrac{n(B\cap C)}{n(S)}=\dfrac{1}{36}$

$P(A\cap C)=\dfrac{n(A\cap C)}{n(S)}=\dfrac{6}{36}=\dfrac1{6}$

(ii) If A and B are independent, then $P(A\cap B)=P(A)P(B)$

$P(A)P(B)=\dfrac{18}{36}.\dfrac{6}{36}=\dfrac{1}{12}=P(A \cap B)$

For B and C:

$P(B)P(C)=\dfrac{6}{36}.\dfrac{6}{36}=\dfrac{1}{36}=P(B \cap C)$

For A and C:

$P(A)P(C)=\dfrac{18}{36}.\dfrac{6}{36}=\dfrac{1}{12}\ne P(A \cap C)$

Hence, A and B, B and C are independent but A and C are not.