# game probability

• Jan 15th 2010, 09:41 PM
thereddevils
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)
• Jan 16th 2010, 02:58 AM
mr fantastic
Quote:

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.
• Jan 16th 2010, 06:25 AM
thereddevils
Quote:

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 ?
• Jan 16th 2010, 06:15 PM
mr fantastic
Quote:

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 ....