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Molly is playing a game that requires her to roll a fair die

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Schools: Columbia Business School, Goizueta, Sloan
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Molly is playing a game that requires her to roll a fair die  [#permalink]

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New post 21 Dec 2011, 13:04
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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?

The solution is this:

P{one roll) +P{two rolls) +P(three rolls) = 1/6 + (5/6*1/6)+(5/6*5/6*1/6) = 91/216

I cannot figure out why in the second and third parenthesis we have 5/6 or 5/6*5/6 respectively. I would put 1/6 instead of 5/6 but I know it would be wrong. Anyone can describe me this in words? Thanks.

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Re: Dice  [#permalink]

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New post 21 Dec 2011, 13:15
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This is a classic example of "OR" and "AND" combination probability.
For a single roll of dice - Prob of getting a "1" is 1/6 and Prob of not getting a 1 is 5/6

Now the probability in question is "She gets a "1" in less than 3 rolls of dice" can be translated into
She gets the "1" in the first roll OR
She gets the "1" in the 2nd roll OR
She gets the "1" in the 3rd roll

Now, consider the first scenario - She gets "1" in the first roll - Probability= 1/6
Second scenario: She doesnt get the "1" in first roll AND gets the "1" in the second roll - Probabilty (5/6) * (1/6)
Third Scenario: She doesnt get "1" in first and second rolls AND gets it in the third roll-
Probability (5/6) * (5/6) * (1/6)

Since all three scenarios are independent of each other we add the 3 results
1/6 + [5/6 *1/6] + [ 5/6 * 5/6 * 1/6]

Hope that helps.
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Re: Dice  [#permalink]

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New post 21 Dec 2011, 13:28
iamgame wrote:
This is a classic example of "OR" and "AND" combination probability.
For a single roll of dice - Prob of getting a "1" is 1/6 and Prob of not getting a 1 is 5/6

Now the probability in question is "She gets a "1" in less than 3 rolls of dice" can be translated into
She gets the "1" in the first roll OR
She gets the "1" in the 2nd roll OR
She gets the "1" in the 3rd roll

Now, consider the first scenario - She gets "1" in the first roll - Probability= 1/6
Second scenario: She doesnt get the "1" in first roll AND gets the "1" in the second roll - Probabilty (5/6) * (1/6)
Third Scenario: She doesnt get "1" in first and second rolls AND gets it in the third roll-
Probability (5/6) * (5/6) * (1/6)

Since all three scenarios are independent of each other we add the 3 results
1/6 + [5/6 *1/6] + [ 5/6 * 5/6 * 1/6]

Hope that helps.


yes, it makes perfect sense. thanks!
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Re: Molly is playing a game that requires her to roll a fair die  [#permalink]

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New post 17 Oct 2018, 20:17
iamgame wrote:
This is a classic example of "OR" and "AND" combination probability.
For a single roll of dice - Prob of getting a "1" is 1/6 and Prob of not getting a 1 is 5/6

Now the probability in question is "She gets a "1" in less than 3 rolls of dice" can be translated into
She gets the "1" in the first roll OR
She gets the "1" in the 2nd roll OR
She gets the "1" in the 3rd roll

Now, consider the first scenario - She gets "1" in the first roll - Probability= 1/6
Second scenario: She doesnt get the "1" in first roll AND gets the "1" in the second roll - Probabilty (5/6) * (1/6)
Third Scenario: She doesnt get "1" in first and second rolls AND gets it in the third roll-
Probability (5/6) * (5/6) * (1/6)

Since all three scenarios are independent of each other we add the 3 results
1/6 + [5/6 *1/6] + [ 5/6 * 5/6 * 1/6]

Hope that helps.


Hi!

Quick question for anyone who may see this, I noticed iamgame said that because the three scenarios are independent, they're added.. however - aren't independent scenarios multiplied?

Look forward to a response, thanks!
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Re: Molly is playing a game that requires her to roll a fair die &nbs [#permalink] 17 Oct 2018, 20:17
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Molly is playing a game that requires her to roll a fair die

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