A person has five tickets of a lucky draw for which a total of 12

Question

A person has five tickets of a lucky draw for which a total of 12 tickets were sold and exactly six prizes are to be given. The probability that the person will win at least one prize is

(A)  \frac{61}{32}

(B) \frac{131}{132}

(C) \frac{31}{132}

(D) \frac{11}{12}

(E) \frac{13}{17}

GMATBusters · Answer

P(At least one prize) = 1- P(No Prize)
= 1- (6C5)/(12C5)
= 1- (6.5.4.3.2.1)/(12.11.10.9.8)
=1-1/132
= 131/132

Answer = B

devavrat · Answer

gmatbusters 
How is 6C5 / 12C5 the probability of winning no ticket?
Can you please explain in detail

GMATBusters · Answer

Out of 12 tickets, 6 tickets are lucky tickets and other 6 are unlucky tickets 
If the person gets all 5 tickets from this 6- unlucky tickets, he will not earn prize. 
The no of ways of selecting 5 tickets out of 6-unlucky tickets = 6C5.

ScottTargetTestPrep · Answer

Note the following formula:

\Rightarrow  P(at least one prize) + P(no prizes) = 1

Thus, if we can determine P(no prizes), we can simply subtract it from 1 to calculate P(at least one prize).

To determine P(no prizes), notice that 6 tickets to earn a prize can be selected from among 12 tickets in 12C6 ways.

\Rightarrow  12C6

\Rightarrow  12!/(6! * 6!)

\Rightarrow  (12 * 11 * 10 * 9 * 8 * 7 * 6!)/(6 * 5 * 4 * 3 * 2 * 6!)

\Rightarrow  2 * 11 * 2 * 3 * 7

If none of the 5 tickets that the person bought earned a prize, then all 6 prizes were chosen from among the remaining 7 tickets. This choice can be made in 7C6 = 7 ways. Thus:

\Rightarrow  P(no prizes) = 7/(2 * 11 * 2 * 3 * 7)

\Rightarrow  1/(2 * 11 * 2 * 3)

\Rightarrow  1/132

Since P(no prizes) = 1/132, it follows that P(at least one prize) = 1 - 1/132 = 131/132.

Answer: B

Jezza · Answer

This is my favorite type of question.
For questions of "whats the probability of getting at least 1 xxx", you always follow the same method: you find the chance of getting 0 xxx, then subtract by 1.

Now for this question, we have 6 tickets with prize, and the other 6 without prize. The combination of getting   and   is 6C0 * 6C5, this is read as "select 0 from the 6 with prize, and select 5 from the 6 without prize", this will give you 6 combinations.

Now we need to find out how many combinations out there in total for randomly selecting 5 tickets out of 12, this is straight 12C5 = 132*6

Notice that the combination of getting at least 1 ticket with prize = 132 * 6 - 6 = 131*6. The probability as such is: 131/132

ThatDudeKnows · Answer

What's the probability of NOT winning?
Don't win on the first drawing: 7/12
Don't win on the second drawing: 6/11
Don't win on the third drawing: 5/10
Don't win on the fourth drawing: 4/9
Don't win on the fifth drawing: 3/8
Don't win on the sixth drawing: 2/7

\frac{7}{12}*\frac{6}{11}*\frac{5}{10}*\frac{4}{9}*\frac{3}{8}*\frac{2}{7} = \frac{1}{132}

What's the probability of YES winning?
 1-\frac{1}{132} = \frac{131}{132}

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A person has five tickets of a lucky draw for which a total of 12

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A person has five tickets of a lucky draw for which a total of 12

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