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805+ (Hard)|   Probability|      
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GMATSkilled
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Mary and Luis were playing a game of rolling 3 fair dice together. If 06 Jun 2018, 17:53
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

41% (02:57) correct 59% (02:44) wrong based on 172 sessions

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Mary and Luis were playing a game of rolling 3 fair dice together. If the product is an odd multiple of 15 Mary wins the game (but if it's an even multiple of 15 Luis wins the game). What is the probability of Mary winning the game ?

A. \(\frac{1}{12}\)

B. \(\frac{1}{15}\)

C. \(\frac{1}{18}\)

D. \(\frac{1}{24}\)

E. \(\frac{5}{24}\)
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C
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EgmatQuantExpert
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Mary and Luis were playing a game of rolling 3 fair dice together. If 06 Jun 2018, 23:32
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Solution



Given:
    • Mary and Luis were playing a game of rolling 3 fair dice together
    • Mary wins if the product is an odd multiple of 15
    • Luis wins if the product is an even multiple of 15

To find:
    • The probability of Mary winning the game

Approach and Working:
As 3 fair dice are rolled together,
    • Total number of possible outcomes = 6 * 6 * 6 = 216

If the product of the 3 outcomes is an odd multiple of 15, then:
    • All the 3 outcomes must be individually odd numbers
    • 2 out of those 3 outcomes must be 3 and 5, in any order

Hence, the possible outcomes of the dice are:
    • 1 – 3 – 5
    • 3 – 3 – 5
    • 5 – 3 – 5

For every selected set, there will be different arrangements
    • Arrangement for 1 – 3 – 5 = 3! = 6
    • Arrangement for 3 – 3 – 5 = \(\frac{3!}{2!}\) = 3
    • Arrangement for 5 – 3 – 5 = \(\frac{3!}{2!}\) = 3

Therefore, total number of outcomes where the product will be an odd multiple of 15 = 6 + 3 + 3 = 12
    • The required probability = \(\frac{12}{216} = \frac{1}{18}\)

Hence, the correct answer is option C.
Answer: C
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PKN
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Mary and Luis were playing a game of rolling 3 fair dice together. If 06 Jun 2018, 22:00
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GMATSkilled
Mary and Luis were playing a game of rolling 3 fair dice together. If the product is an odd multiple of 15 Mary wins the game (but if it's an even multiple of 15 Luis wins the game). What is the probability of Mary winning the game ?

A. \(\frac{1}{12}\)

B. \(\frac{1}{15}\)

C. \(\frac{1}{18}\)

D. \(\frac{1}{24}\)

E. \(\frac{5}{24}\)
Show more


Sample space, S= \(6^3\)

Possible outcomes of the three dice when marry wins(odd multiple of 15):-
1 3 5---------which can occur in 3! different rollings
3 3 5---------which can occur in 3 different rollings
5 3 5---------which can occur in 3 different rollings

So total no of possible rollings when Marry wins= 3!+3+3=12

P(when Marry wins)=P(when the outcome of rollings are odd multiple of 15)=\(\frac{{No. of favorable events}}{{sample space}}\)
=\(\frac{12}{6^3}\)=\(\frac{1}{18}\)

Answer option (C)
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Mary and Luis were playing a game of rolling 3 fair dice together. If 19 Jun 2018, 07:51
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EgmatQuantExpert

Solution



Given:
    • Mary and Luis were playing a game of rolling 3 fair dice together
    • Mary wins if the product is an odd multiple of 15
    • Luis wins if the product is an even multiple of 15

To find:
    • The probability of Mary winning the game

Approach and Working:
As 3 fair dice are rolled together,
    • Total number of possible outcomes = 6 * 6 * 6 = 216

If the product of the 3 outcomes is an odd multiple of 15, then:
    • All the 3 outcomes must be individually odd numbers
    • 2 out of those 3 outcomes must be 3 and 5, in any order

Hence, the possible outcomes of the dice are:
    • 1 – 3 – 5
    • 3 – 3 – 5
    • 5 – 3 – 5

For every selected set, there will be different arrangements
    • Arrangement for 1 – 3 – 5 = 3! = 6
    • Arrangement for 3 – 3 – 5 = \(\frac{3!}{2!}\) = 3
    • Arrangement for 5 – 3 – 5 = \(\frac{3!}{2!}\) = 3

Therefore, total number of outcomes where the product will be an odd multiple of 15 = 6 + 3 + 3 = 12
    • The required probability = \(\frac{12}{216} = \frac{1}{18}\)

Hence, the correct answer is option C.
Answer: C
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EgmatQuantExpert

What rule you used o get the highlighted part?
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Mary and Luis were playing a game of rolling 3 fair dice together. If 20 Jun 2018, 21:32
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hisho
EgmatQuantExpert


• Arrangement for 3 – 3 – 5 = \(\frac{3!}{2!}\) = 3
• Arrangement for 5 – 3 – 5 = \(\frac{3!}{2!}\) = 3

EgmatQuantExpert

What rule you used o get the highlighted part?
Show more
Hey hisho,
The possible number of arrangements, when you have n things and out of those n things r things are identical, is calculated by \(\frac{n!}{r!}\)
This property is used to calculate the indicated highlighted portion
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Mary and Luis were playing a game of rolling 3 fair dice together. If 22 Jun 2018, 10:36
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GMATSkilled
Mary and Luis were playing a game of rolling 3 fair dice together. If the product is an odd multiple of 15 Mary wins the game (but if it's an even multiple of 15 Luis wins the game). What is the probability of Mary winning the game ?

A. \(\frac{1}{12}\)

B. \(\frac{1}{15}\)

C. \(\frac{1}{18}\)

D. \(\frac{1}{24}\)

E. \(\frac{5}{24}\)
Show more

For Mary to win, the throw of the three dice should be {3,5,1} or {3,5,3} or {3,5,5}

Hence the probability is = (1/6*1/6*1/6*3!) + (1/6*1/6*1/6*3!/2!) + (1/6*1/6*1/6*3!/2!) = 1/36 + 1/72 + 1/72 = 1/18

Answer C.

Thanks,
GyM
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Mary and Luis were playing a game of rolling 3 fair dice together. If 30 Oct 2024, 21:22
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Why does order matter in this question? We are not told that there is anything different about each die. By logic, all the dice are the same so shouldn't 1,3,5 or 1,5,3 or 3,1,5 or 3,5,1 or 5,1,3 or 5,3,1 count as the same event. Is it the case that in dice questions, there is an implicit order of ranking similar to questions relating to ordering chairs? I presume that this mechanism would not apply to coins but would it to cards? I request your clarification. Thank you!

EgmatQuantExpert

Solution



Given:
    • Mary and Luis were playing a game of rolling 3 fair dice together
    • Mary wins if the product is an odd multiple of 15
    • Luis wins if the product is an even multiple of 15

To find:
    • The probability of Mary winning the game

Approach and Working:
As 3 fair dice are rolled together,
    • Total number of possible outcomes = 6 * 6 * 6 = 216

If the product of the 3 outcomes is an odd multiple of 15, then:
    • All the 3 outcomes must be individually odd numbers
    • 2 out of those 3 outcomes must be 3 and 5, in any order

Hence, the possible outcomes of the dice are:
    • 1 – 3 – 5
    • 3 – 3 – 5
    • 5 – 3 – 5

For every selected set, there will be different arrangements
    • Arrangement for 1 – 3 – 5 = 3! = 6
    • Arrangement for 3 – 3 – 5 = \(\frac{3!}{2!}\) = 3
    • Arrangement for 5 – 3 – 5 = \(\frac{3!}{2!}\) = 3

Therefore, total number of outcomes where the product will be an odd multiple of 15 = 6 + 3 + 3 = 12
    • The required probability = \(\frac{12}{216} = \frac{1}{18}\)

Hence, the correct answer is option C.
Answer: C
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sam3423
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Mary and Luis were playing a game of rolling 3 fair dice together. If 14 Nov 2024, 05:26
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1. Why make arrangements at all?
2. Also why make arrangements only in the successful cases [ numerator ] and not in the total spaces = 6*6*6 [ This would have a lot of duplicate cases too]
Bunuel Can you help here.
EgmatQuantExpert

Solution



Given:

• Mary and Luis were playing a game of rolling 3 fair dice together
• Mary wins if the product is an odd multiple of 15
• Luis wins if the product is an even multiple of 15

To find:

• The probability of Mary winning the game

Approach and Working:
As 3 fair dice are rolled together,

• Total number of possible outcomes = 6 * 6 * 6 = 216

If the product of the 3 outcomes is an odd multiple of 15, then:

• All the 3 outcomes must be individually odd numbers
• 2 out of those 3 outcomes must be 3 and 5, in any order

Hence, the possible outcomes of the dice are:

• 1 – 3 – 5
• 3 – 3 – 5
• 5 – 3 – 5

For every selected set, there will be different arrangements

• Arrangement for 1 – 3 – 5 = 3! = 6
• Arrangement for 3 – 3 – 5 = \(\frac{3!}{2!}\) = 3
• Arrangement for 5 – 3 – 5 = \(\frac{3!}{2!}\) = 3

Therefore, total number of outcomes where the product will be an odd multiple of 15 = 6 + 3 + 3 = 12

• The required probability = \(\frac{12}{216} = \frac{1}{18}\)

Hence, the correct answer is option C.
Answer: C
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Mary and Luis were playing a game of rolling 3 fair dice together. If 14 Nov 2024, 07:48
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sam3423
1. Why make arrangements at all?
2. Also why make arrangements only in the successful cases [ numerator ] and not in the total spaces = 6*6*6 [ This would have a lot of duplicate cases too]
Bunuel Can you help here.
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The dice are different, so getting 1 on die 1, 3 on die 2, and 5 on die 3 is different from, say, getting 5 on die 1, 3 on die 2, and 1 on die 3. This means you need to account for all different arrangements of getting 1, 3, and 5. The same applies to the other cases.
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