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Diana and Apollo each roll a standard die obtaining a number at random

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Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   11 Mar 2019, 00:42
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Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana’s number is larger than Apollo’s number?

A .\(\frac{1}{3}\)

B. \(\frac{5}{12}\)

C. \(\frac{4}{9}\)

D. \(\frac{17}{36}\)

5. \(\frac{1}{2}\)
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B
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   11 Mar 2019, 10:23
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Total number of cases is 6*6 = 36
Cases favoring Diana = 2; Apollo = 1. 1 case
Diana = 3; Apollo = 2 or 1. 2 cases
Diana = 4; Apollo = 3 or 2 or 1. 3 cases
Diana = 5; Apollo = 4 or 3 or 2 or 1. 4 cases
Diana = 6; Apollo = 5 or 4 or 3 or 2 or 1. 5 cases.
Total cases favoring Diana = 15.

Probability = 15/36 = 5/12.

B is the answer.
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   11 Mar 2019, 22:30
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Solution



Given:
    • Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6.

To find:
    • The probability that Diana’s number is larger than Apollo’s number.

Approach and Working:
For a single fair dice, there will be 6 equally likely outcomes.
    • Therefore, for 2 such dice, total possible outcomes = 6 x 6 = 36

Out of the 36 outcomes,
    • In 6 outcomes both the dice will show same number.
From the remaining (36 – 6) = 30 outcomes, half will have first outcome greater than the second, and the other half will have second outcome greater than the first.
    • Hence, number of possible cases where Diana’s outcome greater than Apollo’s outcome = 15
    • Therefore, required probability = \(\frac{15}{36}\)

Hence, the correct answer is option B.

Answer: B

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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   11 Mar 2019, 23:18
Ok so let’s denote number obtained by both using a pair. So if Diane gets a 1 and Apollo gets a 2 we represent it as (1,2). Thus the number of cases where Dianne gets a number greater than Apollo is 15. (Rather than waste time getting some convoluted logic, just enumerate all pairs of numbers between 1 and 6 (inclusive) where one number is greater than another.These are (1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6). Total number of possible pairs will be 6X6=36.
So ans = 15/36 = 5/12 = B

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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   11 Mar 2019, 23:53
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TyrionLannister wrote:
Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana’s number is larger than Apollo’s number?

A .\(\frac{1}{3}\)

B. \(\frac{5}{12}\)

C. \(\frac{4}{9}\)

D. \(\frac{17}{36}\)

5. \(\frac{1}{2}\)


There are 6 outcomes for Diana and 6 outcomes for Apollo. So in all there are 6*6 = 36 combinations of outcomes.
Out of these 6 outcomes will be the same for both Diana and Apollo (1, 1), (2, 2) ... (6, 6)
Of the remaining 30 combinations, in half of them Diana's outcome would be higher and in the other half, Apollo's outcome will be higher.

Probability of Diana’s number being larger than Apollo’s number = 15/36 = 5/12

Answer (B)
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   14 Mar 2019, 02:01
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TyrionLannister wrote:
Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana’s number is larger than Apollo’s number?

A .\(\frac{1}{3}\)

B. \(\frac{5}{12}\)

C. \(\frac{4}{9}\)

D. \(\frac{17}{36}\)

5. \(\frac{1}{2}\)


total P = 6*6 = 36
and for each roll Diana getting greater no than Apollo would be
D=2, A = 1
D=3, A= 1,2
D=4,A = 1,2,3
D=5, A = 1,2,3,4
D=6, A = 1,2,3,4,5
total A = 15 digits
P = 15/36 ; 5/12
IMO B
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   28 Apr 2021, 17:55
Why is my approach wrong?

A = 1 <--What Apollo rolls, then D = 5/6 <--- Probability that she gets a higher number
A = 2, D = 2/3
A = 3, D = 1/2
A = 4, D = 1/3
A = 5, D = 1/6
A = 6, D = 0

5/6 + 2/3 + 1/2 + 1/3 + 1/6

???
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink] New post   28 Apr 2021, 19:12
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CEdward wrote:
Why is my approach wrong?

A = 1 <--What Apollo rolls, then D = 5/6 <--- Probability that she gets a higher number
A = 2, D = 2/3
A = 3, D = 1/2
A = 4, D = 1/3
A = 5, D = 1/6
A = 6, D = 0

5/6 + 2/3 + 1/2 + 1/3 + 1/6

???


A=1, D could be 6,5,4,3,2 --->5 ways
A=2, D could be 6,5,4,3 ---> 4 ways
A=3, D could be 6,5,4 ---> 3 ways
A=4, D could be 6,5 ---> 2 ways
A=5, D could be 6 ---> 1 way

then you need to sum up these possible possible ways: 5+4+3+2+1=15 ways

these two persons roll the die seperately, thus the total ways should be 6*6=36

so the probability that the number of Diana is larger than that of Apollo is 15/36=5/12
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Re: Diana and Apollo each roll a standard die obtaining a number at random [#permalink]
28 Apr 2021, 19:12
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