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by rahul.s » Mon Feb 08, 2010 4:55 am
molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. what is the probability that molly will roll the die less than four times before stopping?

[spoiler]OA: 91/216[/spoiler]
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by shashank.ism » Mon Feb 08, 2010 5:16 am
rahul.s wrote:molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. what is the probability that molly will roll the die less than four times before stopping?

[spoiler]OA: 91/216[/spoiler]
S= 6^3 =216
E = 1C1+5C1*6C1+5C1*5C1*1C1= 1+5+25
P = 31/216[spoiler]31/216[/spoiler]
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by harsh.champ » Mon Feb 08, 2010 5:20 am
shashank.ism wrote:
rahul.s wrote:molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. what is the probability that molly will roll the die less than four times before stopping?

[spoiler]OA: 91/216[/spoiler]
S= 6^3 =216
E = 1C1+5C1*6C1+5C1*5C1*1C1= 1+5+25
P = 31/216[spoiler]31/216[/spoiler]
Now,molly will roll the die less than 4 times if she gets 1 on the 1st,2nd or the 3rd time .

P = 1/6 + (5/6 x 1/6) + (5/6 x 5/6 x 1/6)
=[spoiler] (36 + 30 + 25)/216 =91/216[/spoiler]

Hey Shashank,
I think you forgot to multiply the 1st and 2nd term with 36 and 6 respectively.

Just check with your problem approach.
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