JEE Main 2023 — Mathematics Probability & Statistics

Given, $A$A: No. on ${1}^{\text{st }}$1st die $<$< No. on ${2}^{\text{nd }}$2nd  die

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

$$\therefore n(A)=15$$∴n(A)=15

$B$B: No. on ${1}^{\text{st }}$1st  die $=$=even & No. of ${2}^{\text{nd }}$2nd  die $=$= odd

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

$$\therefore n(B)=9$$∴n(B)=9

$C:$C: No. on ${1}^{\text{st }}$1st  die $=$=odd & No. on ${2}^{\text{nd }}$2nd  die $=$=even

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

$$\therefore n(C)=9$$∴n(C)=9

Now,

$$n(A\cap B)=3,n(A\cap C)=6,n(B\cap C)=0$$n(A∩B)=3,n(A∩C)=6,n(B∩C)=0

$$n(A\cap B\cap C)=0$$n(A∩B∩C)=0

Since $(4,5)\in A$(4,5)∈A and $(4,5)\in B$(4,5)∈B

$\therefore$∴ $A$A and $B$B are not exclusive events

Now,

$$n((A\cup B)\cap C)=n(A\cap C)+n(B\cap C)-n(A\cap B\cap C)$$n((A∪B)∩C)=n(A∩C)+n(B∩C)−n(A∩B∩C)

$$=6$$=6

Since, $P(B)=\frac{9}{36},P(C)=\frac{9}{36},P(B\cap C)=0$P(B)=369​,P(C)=369​,P(B∩C)=0

$$\Rightarrow P(B).P(C)\neq P(B\cap C)$$⇒P(B).P(C)=P(B∩C)

$\therefore B$∴B and $C$C are not independent