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Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?
(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.
(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.
11 Nov 2013, 01:58
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Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?
(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5. Notice that this statement tells us that Rachel throws the die twice. Also, we are told that \(P(x)=\frac{1}{5}\), thus \(P(not \ x)=1-\frac{1}{5}=\frac{4}{5}\). Therefor \(P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}\). Sufficient.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. \(P(at \ least \ one \ x)=1-P(no \ x's)=\frac{61}{125}\) --> \(P(no \ x's)=\frac{64}{125}\). Since \(P(no \ x's)\) for n throws also equals to \((\frac{x-1}{x})^n\), then \((\frac{x-1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3\) --> \(x=5\) and \(n=3\) is the only integer solution. Thus Rachel thrown the die n-1=2 times and x=5 --> \(P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}\). Sufficient.
Answer: D.
05 Nov 2014, 14:38
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statement 1 :
Probability of getting x is 1/5 . So probability of not getting an X is 4/5. Die is throw twice.
To get at least one x
-> first throw get x & second throw not get x
-> first throw not get X & second throw get X
-> get x on both rolls.
So probability of getting at least one X is 2*[1/5*4/5] + [1/5*1/5] = 9/25 . Hence statement 1 sufficient. [We need not calculate..we can stop with knowing that calculation can be done]
Statement 2:
Rachel had thrown the die for some n times , if she had thrown it for n+1 times , the probability of getting at least one x is 61/125.
When at least once is given, we can always say 1 - P (none) = P (at least once)
1 - P (none) = 61/125
P (none) = 1 - [61/125] = 64/ 125
Each throw is an independent event and hence the probability of not getting X in one throw will be the same as the probability of not getting X in the next throw and all subsequent throws if any. So, P [none] should be a perfect square only if the same fraction is multiplied two or more times.
So P[none ] = [4/5] * [4/5] * [4/5]
i.e. Rachel does three throws and gets X in none.
Now (n+1) is three throws, so n is two throws & probability of not getting X is 4/5, so probability of getting X is 1/5. Now the problem looks similar to Statement and we will get the same answer 9/25. [But we need even consider all this, we can stop with P [none] = [4/5]^3 knowing that it can be done after that]
Hence statement alone is also sufficient.
Thus Ans -> D [Note: Do not get confused with a fact that a regular die has six faces, here consider that the die Rachel uses has only 5 faces...it becomes easy].
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