Question

A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses.​

Answer 1

Answer:

If a die is thrown three times, then the number of elements in the sample space will be 6×6×6=216

E={(1,1,4),(1,2,4),…(1,6,4)(2,1,4),(2,2,4),…(2,6,4)(3,1,4),(3,2,4),…(3,6,4)(4,1,4),(4,2,4),…(4,6,4)(5,1,4),(5,2,4),…(5,6,4)(6,1,4),(6,2,4),…(6,6,4)}

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

∴E∩F={(6,5,4)}

P(F)=

216

6

and P(E∩F)=

216

1

∴P(E∣F)=

P(F)

P(E∩F)

=

216

6

216

1

=

6

1

=0.16

Step-by-step explanation:

.$\huge\color{yellow}{@ᴍɪɴɴɪᴇ ★}$

Answer 2

Step-by-step explanation:

Let E: 4 appears on the third toss

F: 6 and 5 appear respectively on the first two tosses.

P(E)=

6

1

P(F)=

6

1

×

6

1

=

36

1

Now,

P(E∩F)=

216

1

Therefore, required probability , P(E∣F)=

P(F)

P(E∩F)

=

6

1