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Bunuel
If Enid and Topanga each roll a single ten-sided die (which has sides numbered 1 through 10), what is the probability that Enid will roll a larger number than Topanga does?

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

B. \(\frac{9}{20}\)

C. \(\frac{49}{100}\)

D. \(\frac{99}{200}\)

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

Official solution from Veritas Prep.

This problem combines two different ideas central to higher-level probability questions: pairs probability, and bad outcome counting.

We begin by noting that there are really three types of outcomes here: Enid rolls a larger number, Topanga rolls a larger number, or the two of them tie.

To calculate the probability of a tie, we use pairs probability; take the die rolls one at a time and ask, at each step, “am I still happy?” Since we’re going for a pair, any first roll will make us happy – every number is the first half of some possible pair. When the time comes for the second roll, on the other hand, we only have a \(\frac{1}{10}\) chance of being happy – we have to match whatever number was rolled before. So overall there is a \(1∗\frac{1}{10}=\frac{1}{10}\) chance of a tie, and a \(1-\frac{1}{10}=\frac{9}{10}\) chance that one die roll will be larger than the other.

To calculate the probability that Enid rolls a larger number, observe that it is equally as likely that Enid will win as it is that Topanga will win; there’s perfect symmetry between the two possibilities. So the probability that Enid will roll a larger number is simply half of the probability that someone will roll a higher number. That is,

\(P(Enid)=\frac{1}{2}∗\frac{9}{10}=\frac{9}{20}\). This is \(B\).
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There are three possibilities
Enid=Topanga Enid>Topanga Enid<Topanga

Probability of Enid=Topanga is 1/10
so, probablity of Enid>Topanga Enid<Topanga is 1-1/10 = 9/10
1/2 of this will be greater and 1/2 will be lower.
So, probability that Enid will roll a larger number than Topanga is 9/10 * 1/2 = 9/20 B
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this is a clear 700!
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Not the hardest but here is one of those high level problems where you can bank some time for later ones by truly understanding the concept.

First I calculated the number of potential outcomes which is 10 * 10 = 100;

Next, figure out how many different ways in which a tie can occur which is just the number of possibilities where Enid's number equals Topanga which would be the total number of distinct outcomes (1, 2, 3, ... 10) which is 10;

Finally, its important to understand that without any special rules, outside of ties, the remaining possibilities are either wins or losses for Enid. Since they have equal probability of rolling each number, they are just as likely to win as they are to lose, one-half of the remaining outcomes represents the probability of a loss (as well as a win) for each of them:

100 - 10 = 90 possible non-tie outcomes;

90 / 2 = 45 possible outcomes where Enid wins;

45/100 --> reduce by 5 and you end up with 9/20, answer choice B;
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Bunuel
If Enid and Topanga each roll a single ten-sided die (which has sides numbered 1 through 10), what is the probability that Enid will roll a larger number than Topanga does?

A. 2/5
B. 9/20
C. 49/100
D. 99/200
E. 1/2


Total # of outcomes when 2, 10 sided dice are rolled = 10 * 10 = 100

Favorable Condition, Enid rolls a larger # than Topanga.
Hence we have, below possibilities,
Enid = 10 Topanga = 9, 8, 7,....1 = 9 possibilities
Enid = 9 Topanga = 8,7,6,......1 = 8 possibilities

.
.
.
Enid = 2 Topanga = 1 = 1 possibility

Total # of possibilities = 9+8+7+...+1 = 45

Required Probability = 45/100 = 9/20

Answer B.

Thanks,
GyM
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    Solution



    Given
      • Enid and Topanga each roll a single ten-sided dice.

    To find
      • The probability that Enid will roll a larger number than Topanga.

    Approach and Working out
      • Total possibility = 10 * 10 = 100
      • Number of cases when Enid will roll a larger number than Topanga.

        o Enid get 10, Topanga can get 9 numbers (1, 2, 3, 4, 5, 6, 7, 8, 9) on his dice.
        o Enid get 9, Topanga can get 8 numbers (1, 2, 3, 4, 5, 6, 7, 8) on his dice.
        o Enid - 8, Topanga - 7
        o Enid - 7, Topanga - 6
        o Enid - 6, Topanga - 5
        o Enid - 5, Topanga - 4
        o Enid - 4, Topanga - 3
        o Enid - 3, Topanga - 2
        o Enid - 2, Topanga - 1
        o Enid - 1, Topanga - 0

    So, total 1+2+3+4+5+6+7+8+9 = 45 cases.

      • Hence, probability = 45/100 = 9/20

    Hence, option C is the correct answer.

    Correct Answer: Option C
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If Enid and Topanga each roll a single ten-sided die (which has sides numbered 1 through 10)

As we are rolling two dice => Number of cases = \(10^2\) = 100

What is the probability that Enid will roll a larger number than Topanga does?

Out of these 100 cases following can be the outcome:

  • Both rolls are equal, number of cases = 10 [(1,1), (2,2), (3,3)....(10,10)]
    • Remaining cases = 100 - 10 = 90
  • First roll > Second Roll = 90/2 = 45 cases
  • Second Roll > First Roll = 90/2 = 45 cases

=> Probability that Enid will roll a larger number than Topanga = 45/100 = 9/20

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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