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# Question of the Week- 20 (A biased dice is such that any odd number..)

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Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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26 Oct 2018, 04:51
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85% (hard)

Question Stats:

42% (02:32) correct 58% (02:29) wrong based on 69 sessions

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Question of the Week #20

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

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Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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26 Oct 2018, 05:17
1
EgmatQuantExpert wrote:

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

If Probability of even Number = x
then Probability of odd number = 3x

But, Probability of even Number on dice + Probability of odd Number on dice = 1

i.e. x+3x = 1
i.e. x = 1/4

Sum may be 10 in following ways
(4,6) (5,5) (6,4)

Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144

Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144)

EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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29 Oct 2018, 20:18
GMATinsight
Hi,
I am not able to understand why you multiplied 1/3 .
ex: probability of 4 = 1/4*1/3
Thanks
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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29 Oct 2018, 20:27
4
HAPPYatHARVARD wrote:
GMATinsight
Hi,
I am not able to understand why you multiplied 1/3 .
ex: probability of 4 = 1/4*1/3
Thanks

HAPPYatHARVARD

probability of 4 = Probability of getting even number * probability of getting 4 out of 3 even numbers {2,4,6}
Now probability of getting an even number on this dice = 1/4
Probability of getting 4 out of three even numbers {2,4,6} = 1/3
i.e. Probability of getting 4 on unbiased dice = 1/4*1/3

I hope this helps
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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29 Oct 2018, 20:38
Please elaborate why did you multiply all probabilities with 1/3?

GMATinsight wrote:
EgmatQuantExpert wrote:

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

If Probability of even Number = x
then Probability of odd number = 3x

But, Probability of even Number on dice + Probability of odd Number on dice = 1

i.e. x+3x = 1
i.e. x = 1/4

Sum may be 10 in following ways
(4,6) (5,5) (6,4)

Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144

Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144)

EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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29 Oct 2018, 20:44
1
gvij2017 wrote:
Please elaborate why did you multiply all probabilities with 1/3?

GMATinsight wrote:
EgmatQuantExpert wrote:

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

If Probability of even Number = x
then Probability of odd number = 3x

But, Probability of even Number on dice + Probability of odd Number on dice = 1

i.e. x+3x = 1
i.e. x = 1/4

Sum may be 10 in following ways
(4,6) (5,5) (6,4)

Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144

Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144

Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144)

EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well

gvij2017

Rationale for using the probabilities is as follows:

Probability of getting an even number on dice = 1/4
Probability of getting an Odd number on dice = 3/4

Probability of getting a specific even number e.g. 2 out of three even numbers {2,4,6} = 1/3
Probability of getting a specific odd number e.g. 3 out of three odd numbers {1,3,5} = 1/3

In each case either we calculate probability of a specific even number or an odd number therefore in each case we have used 1/3 for that specific even number or odd number.

I hope this helps!!!
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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01 Nov 2018, 03:17
1
1

Solution

Given:
• In a biased dice, any odd number appears thrice as frequently as any even number

To find:
• The probability that the sum of the numbers obtained upon rolling the dice twice is 10

Approach and Working:
• Let us assume that the frequency of getting any even number, {2, 4, 6} is x
o Then, the frequency of getting any odd number, {1, 3, 5} will be 3x

• Implies,
o The probability of getting any even number = $$\frac{x}{(x + 3x + x + 3x + x + 3x)} = \frac{1}{12}$$
o And, the probability of getting any odd number = $$\frac{3x}{(x + 3x + x + 3x + x + 3x)} = \frac{3}{12} = \frac{1}{4}$$

• For the sum of two outcomes to be 10, the outcomes must be either (4, 6) or (5, 5) or (6, 4)
• Therefore, P(sum = 10) = P(4, 6) + P(5, 5) + P(6, 4) = $$(\frac{1}{12} * \frac{1}{12}) + (\frac{1}{4} * \frac{1}{4}) + (\frac{1}{12} * \frac{1}{12}) = \frac{11}{144}$$

Hence the correct answer is Option C.

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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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02 Nov 2018, 10:55
why not probability of 4 = Probability of getting even number * probability of getting 4 out of six numbers.

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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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02 Nov 2018, 20:52
1
HAPPYatHARVARD wrote:
why not probability of 4 = Probability of getting even number * probability of getting 4 out of six numbers.

GMATinsight

HAPPYatHARVARD

Once you have included the probability of getting even number then the total choices that you are left with only three numbers {2, 4, 6} hence the following probability can NOT be 'probability of getting 4 out of six numbers', It has to be the probability of 4 out of three even numbers.
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Re: Question of the Week- 20 (A biased dice is such that any odd number..)  [#permalink]

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24 Jun 2019, 05:48
EgmatQuantExpert wrote:
Question of the Week #20

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

11/144
Let us assume that the frequency of getting any even number, {2, 4, 6} is x
Then, the frequency of getting any odd number, {1, 3, 5} will be 3x
Implies,
The probability of getting any even number = X/(3X+(3*3X))=1/12
SIMILARLY ODD NUMBER=2/12
TOTAL COMBINATIONS OF GETTING 10 ARE 4+6,5+5,6+4
Re: Question of the Week- 20 (A biased dice is such that any odd number..)   [#permalink] 24 Jun 2019, 05:48
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