A and B throw a dice alternatively till one of them gets a ‘6’and wins the game. Find their respective probabilities of winning,if A starts first??

P(A w ins)= P(S)+ P(FFS)+ P(FFFFS)=given below

 and so on.

step 2 =    1/6/1-25/36 = 6/11 please tell me fast.

how step 2 came

Hi Rahul
Here is the answer to your query.
 
Given A and B throw a dice simultaneously till one thus gets a 6 and wins this game.
Now, consider winning of A
Let S and F be the following events.
S: obtaining a six by A
F: not obtaining a six by B
 
A can win the following way:
(i) He obtains a six in the first trial
(ii) A fails to obtain a six, B fails to obtain and then A obtains a six
(iii) A fails, then B fails, then A fails, then B fails to obtain and finally A obtains a six and so on
 
Obtaining a six by A and not obtaining a six by B are independent events
 
 
 
Hope! This will help you.
Cheers!!!

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