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Sierra, Randolph, and Conrad all submitted a different number of ticke

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Sierra, Randolph, and Conrad all submitted a different number of ticke  [#permalink]

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New post 07 Apr 2016, 08:43
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A
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C
D
E

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Sierra, Randolph, and Conrad all submitted a different number of tickets for a raffle, and only one of them will win. What is the probability of Conrad's winning the raffle?

(1) The probability that Conrad will win is twice the probability that Sierra will win.
(2) The probability that Sierra will win is 1/3 the probability that Randolph will win.

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Re: Sierra, Randolph, and Conrad all submitted a different number of ticke  [#permalink]

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New post 07 Apr 2016, 12:34
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Is it C? p(randall)+p(conrad)+p(sierra)=1 because one of them will win, as the problem says. So for the first statement 2p(s)+p(s)+p(r)=1.. that's not enough. 2/3p(r)+1/3p(r)+p(r)=1 p(r)=1/2 p(s)=1/6 and p(c)=2*1/6=1/3
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Re: Sierra, Randolph, and Conrad all submitted a different number of ticke  [#permalink]

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New post 24 Nov 2019, 10:51
Bunuel wrote:
Sierra, Randolph, and Conrad all submitted a different number of tickets for a raffle, and only one of them will win. What is the probability of Conrad's winning the raffle?

(1) The probability that Conrad will win is twice the probability that Sierra will win.
(2) The probability that Sierra will win is 1/3 the probability that Randolph will win.



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Assume s, r and c are probabilities for Sierra, Randolph and Conrad.
We have conditions s + r + c = 1 and s≥0, r≥0, c≥0.

Since we have 3 variables (s, r and c) and 1 equation, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

We have c = 2s and s = (1/3)r. Then c = (2/3)r
s + r + c = (1/3)r + r + (2/3)r = 2r = 1.
Then we have r = 1/2, s = 1/6 and c = 1/3.

Since both conditions together yield a unique solution, they are sufficient.

Since this question is a probability question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
We have c = 2s.
We can consider both cases of r = 1/2, s = 1/6, c = 1/3 and r = 5/8, s = 1/8, c = 1/4.
There are a lot of possibilities.

Since condition 1) does not yield a unique solution, it is not sufficient


Condition 2)
We have s = (1/3)r.
We can consider both cases of r = 1/2, s = 1/6, c = 1/3 and r = 1/3, s = 1/9, c = 5/9.
There are a lot of possibilities.

Since condition 2) does not yield a unique solution, it is not sufficient

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Sierra, Randolph, and Conrad all submitted a different number of ticke   [#permalink] 24 Nov 2019, 10:51
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Sierra, Randolph, and Conrad all submitted a different number of ticke

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