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1994 hsc question 10 (1 Viewer)

1zephyr

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david has invented a game for one person. he throws two ordinary dice repeatedly until the sum of the two numbers shown is either 7 or 9. if the sum is 9, David wins. If the sum is 7, David loses. If the sum is any other number, he continues to throw until until it is 7 or 9

iii) What is the probability David wins on his first, second, or third throw? Leave your answer in unsimplified form

answer is (13/12)^2 x 1/9
why is that???? cheers brahhhhhh
 

seventhroot

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draw one of those tables; get the values and then it becomes a standard question

what a sad game lol
 

1zephyr

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i did brah but i still dont get why you have to multiply by 13/18 which is the odd of not winning on the first
 

1zephyr

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nevermind ya sick kunti just figured it out. fuck ur a beast
 

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