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In a contest, a bowl contains 10 keys, one of which will open a treasu
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22 Apr 2015, 03:14
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63% (01:07) correct 38% (01:08) wrong based on 184 sessions
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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23 Apr 2015, 04:20
jayanthjanardhan wrote: Can i please have the solution to this? Probability of having the right key = Favorable Cases/Total Cases = (Cases in which the right key will be picked)/(No of ways in which you can pick two keys) No of ways in which you can pick two keys out of 10 = 10C2 = 10*9/2 = 45 No of cases in which the right key will be picked = 1*9C1 (you pick the correct key and one other key) = 9 Probability of winning = 9/45 = 1/5 Answer (C) You can also do the question with arrangement since you need probability. Pick the first key in 10 ways and second in 9 ways. Total = 10*9 = 90 ways Pick the first correct key in 1 way and second key in 9 ways. Total = 1*9 = 9 ways Pick the first key in 9 ways and second correct key in 1 way. Total = 9*1 = 9 ways Probability = (9+9)/90 = 1/5 Another Method is using probabilitiesShe can win the contents by picking the first key correctly or second key correctly. Probability of picking the first key correctly = 1/10 Probability of picking the first key incorrectly and second key correctly = (9/10)*(1/9) = 1/10 Probability of winning = 1/10 + 1/10 = 1/5
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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22 Apr 2015, 03:23
Bunuel wrote: In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?
A. 1/10 B. 1/9 C. 1/5 D. 19/90 E. 2/9
Kudos for a correct solution. The total possibilities existing while withdrawing two keys are: 10C2 = 45 For Anna to win, 9 combinations exist assuming Key 1 to unlock the treasure E.g(1,2), (1,3)....(1,9) P = 9/45 = 1/5 Option C



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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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22 Apr 2015, 22:44
Can i please have the solution to this?



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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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24 Apr 2015, 11:15
Hi All, When dealing with probability questions, there are 2 outcomes that you can calculate: what you WANT to have happen and what you DON'T WANT to have happen. When you add these two outcomes together, the result will equal 1. Here, it's actually pretty easy to calculate the outcome that you DON'T WANT: NOT getting the winning key. Then we can subtract that fraction from 1 to get the probability of what we do WANT... There are 10 keys and 1 of them is the 'winning' key. Probability of NOT getting the winning key on the first grab = 9/10 Probability of NOT getting the winning key on the second grab = 8/9 (9/10)(8/9) = 8/10 = 4/5 So, the probability of NOT getting the winning key = 4/5 The probability that we DO get the winning key = 1  4/5 = 1/5 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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27 Apr 2015, 04:06
Bunuel wrote: In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?
A. 1/10 B. 1/9 C. 1/5 D. 19/90 E. 2/9
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:When Anna reveals her two keys, there are two possible sequences that allow her to win: 1) The first key is the winner, the second is not 2) The first key is not the winner, but the second one is And you can calculate those probabilities: 1) 1/10 * 9/9 = 1/10 2) 9/10 * 1/9 = 1/10 Add those two favorable probabilities and you have 2/10 which reduces to 1/5.
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In a contest, a bowl contains 10 keys, one of which will open a treasu
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05 Jun 2016, 03:44
since the two keys have been drawn simultaneously(and not one by one) and hence the no of cases must be 10 c2 not 10 *9. Now
one key is correct one and second key incorrect one= 1* 9c1
there is no logic of first and second key as the draw was simutaneous. so probability =9c1 /10c2=1/5



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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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05 Jun 2016, 21:02
vishnu440 wrote: since the two keys have been drawn simultaneously(and not one by one) and hence the no of cases must be 10 c2 not 10 *9. Now
one key is correct one and second key incorrect one= 1* 9c1
there is no logic of first and second key as the draw was simutaneous. so probability =9c1 /10c2=1/5 Note that considering first and second key in a valid method even though we talk of simultaneous draws because when you pick out two, you will touch one of them first. They are both drawn out simultaneously but not picked simultaneously. Of course when you consider that first key is A and second key is B, you have to consider the other side too where the first key is B and second key is A. The answer will be the same no matter which method you use  combinations or probability.
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In a contest, a bowl contains 10 keys, one of which will open a treasu
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05 Jun 2016, 21:24
Bunuel wrote: In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?
A. 1/10 B. 1/9 C. 1/5 D. 19/90 E. 2/9
Kudos for a correct solution. There are two ways in which the right key is taken out (1st key is the right one & second wrong one) or (1st key is the wrong one & 2nd one is right) Since it is "or" hence we will add the probabilities. ( \(\frac{1}{10}\) * \(\frac{9}{9}\)) + ( \(\frac{9}{10}\) * \(\frac{1}{9}\)) \(\frac{1}{10}\) = Only one key is the right key. Total options are 10 \(\frac{9}{9}\) = After first key is taken out, there r 9 favorable options (all keys r wrong). Total options are also 9 \(\frac{9}{10}\) = For first key to be the wrong one, favorable options are nine. Total options are 10. \(\frac{1}{9}\) = For second key to be correct one, only one favorable option is there out of available 9 options. \(\frac{9}{90}\) + \(\frac{9}{90}\) \(\frac{18}{90}\) = \(\frac{1}{5}\)C is the answer.



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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu
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21 Sep 2017, 15:02
Bunuel wrote: In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?
A. 1/10 B. 1/9 C. 1/5 D. 19/90 E. 2/9 P(Anna wins) = 1  P(Anna does not win) Let’s determine P(Anna does not win). Note that the probability that the first key doesn’t open the treasure chest is 9/10 and the probability that the second key doesn’t open the treasure chest (given that the first key didn’t work) is 8/9. P(Anna does not win) = 9/10 x 8/9 = 4/5. Thus, the probability that she wins is 1  ⅘ = 1/5. Answer: C
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu &nbs
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