Thread: game probability

A game is played by players taking turns toss a die . If a player gets 2,3,4 or 5 , then it would be his first score and the next player takes over . If the score is 1 or 6 , the player is given an extra toss and his score would be the sum of the two scores obtained . Events A and B are defined as follows .

A=The score is 5,6,7,8 or 9 .

B=The player tosses twice .

Show that p(A)=1/3 . Find P(B) and P(AnB)

Originally Posted by thereddevils

A game is played by players taking turns toss a die . If a player gets 2,3,4 or 5 , then it would be his first score and the next player takes over . If the score is 1 or 6 , the player is given an extra toss and his score would be the sum of the two scores obtained . Events A and B are defined as follows .

A=The score is 5,6,7,8 or 9 .

B=The player tosses twice .

Show that p(A)=1/3 . Find P(B) and P(AnB)

I suggest drawing a tree diagram. By the way, Pr(B) should be easy ....

If you need more help, please show what you could do and say where you're stuck.

Originally Posted by mr fantastic

I suggest drawing a tree diagram. By the way, Pr(B) should be easy ....

If you need more help, please show what you could do and say where you're stuck.

i am not sure how to draw a tree diagram for such questions , i tried counting the possibilities but just to no avail . Can you get me started ?

Originally Posted by thereddevils

i am not sure how to draw a tree diagram for such questions , i tried counting the possibilities but just to no avail . Can you get me started ?

The first two branches are "2,3,4 or 5" and "1, 6". And only one of these branches will have other branches coming from it ....