Consider the word POSSIBILITY. In the arrangement of the letters of the above word let `A` and `B` denote the event that the `2 S's` are together and the `3I's` are together respectively, then

Options

A. `P(A)=P(B)=(3)/(11)`
B. `P(AnnB)=(2)/(165)`
C. `P(AuuB)=(7)/(31)`
D. `P(B//A)=(1)/(15)`

Answer

B, D

Explanation

`(b,d)`In the word ''POSSIBILITY'' there are `2S's`, `3I's` and `6` other different words.
`P(A)=((10!)/(3!))/((11!)/(3!2!))=(2)/(11)`
`P(B)=((9!)/(2!))/((11!)/(3!2!))=(3)/(55)`
`P(AnnB)=(8!)/((11!)/(3!2!))=(2)/(165)`
`P(AuuB)=P(A)+P(B)-P(AnnB)`
`=(2)/(11)+(3)/(55)-(2)/(165)=(30+9-2)/(165)=(37)/(165)`
`P(B//A)=(P(AnnB))/(P(A))=((2)/(165))/((2)/(11))=(1)/(15)`