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Molly is rolling a number cube with faces numbered 1 to 6

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AccipiterQ
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Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   24 Oct 2013, 10:07
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Molly is rolling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?

A) 11/180

B) 125/216

C) 25/216

D) 91/216

E) 27/128
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Bunuel
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   24 Oct 2013, 10:15
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AccipiterQ wrote:
Molly is rolling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?

A) 11/180

B) 125/216

C) 25/216

D) 91/216

E) 27/128


will post OE after a few attempts


The probability that Molly will roll the die less than 4 times before stopping is the sum of the following:

The probability that Molly will roll the die once: P=1/6 (she gets 1 on the first roll);
The probability that Molly will roll the die twice: P=5/6*1/6 (not 1, 1);
The probability that Molly will roll the die thrice: P=5/6*5/6*1/6 (not 1, not 1, 1).

P = 1/6 + 5/6*1/6 + 5/6*5/6*1/6 = 91/216.

Answer: D.
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   04 Nov 2013, 11:54
Bunuel wrote:
AccipiterQ wrote:
Molly is rolling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?

A) 11/180

B) 125/216

C) 25/216

D) 91/216

E) 27/128


will post OE after a few attempts


The probability that Molly will roll the die less than 4 times before stopping is the sum of the following:

The probability that Molly will roll the die once: P=1/6 (she gets 1 on the first roll);
The probability that Molly will roll the die twice: P=5/6*1/6 (not 1, 1);
The probability that Molly will roll the die thrice: P=5/6*5/6*1/6 (not 1, not 1, 1).

P = 1/6 + 5/6*1/6 + 5/6*5/6*1/6 = 91/216.

Answer: D.


I tried looking at it this way:
1-(5/6)^4
what is wrong with this logic?
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   28 Apr 2014, 21:55
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ronr34 wrote:
Bunuel wrote:
AccipiterQ wrote:
Molly is rolling a number cube with faces numbered 1 to 6 repeatedly. When she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?

A) 11/180

B) 125/216

C) 25/216

D) 91/216

E) 27/128


will post OE after a few attempts


The probability that Molly will roll the die less than 4 times before stopping is the sum of the following:

The probability that Molly will roll the die once: P=1/6 (she gets 1 on the first roll);
The probability that Molly will roll the die twice: P=5/6*1/6 (not 1, 1);
The probability that Molly will roll the die thrice: P=5/6*5/6*1/6 (not 1, not 1, 1).

P = 1/6 + 5/6*1/6 + 5/6*5/6*1/6 = 91/216.

Answer: D.


I tried looking at it this way:
1-(5/6)^4
what is wrong with this logic?


A small thing: It should actually be 1 - (5/6)^3

P(Less than 4 times) = 1 - P(4 times or more)

When will we roll the die 4 times or more? When we get something other than 1 in each of the first 3 rolls.

P(4 times or more) = (5/6)*(5/6)*(5/6)

P(Less than 4 times) = 1 - (5/6)*(5/6)*(5/6) = 91/216
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   28 Jun 2019, 14:36
P(Roll 1 on 1st attempt OR Roll 1 on 2nd attempt OR Roll 1 on 3rd attempt)
Roll 1 on 1st = 1/6
Roll 1 on 2nd = 5/6*1/6 = 5/36
Roll 1 on 3rd = 5/6*5/6*1/6 = 25/216

1/6 + 5/36 + 25/216 =
(36 + 30 + 25) / 216 =
91 / 216
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink] New post   27 Feb 2024, 07:21
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Re: Molly is rolling a number cube with faces numbered 1 to 6 [#permalink]
27 Feb 2024, 07:21
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