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 Draw and go (Posted on 2014-10-08)
A class, consisting of g girls and b boys receieved 2k promotional theatre tickets.
Given that b<2k and g<2k, what is the probability that within the group of theatre-goers there will be more girls than boys?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 re(2): Solution or partial solution ... clarification | Comment 3 of 14 |
(In reply to re: Solution or partial solution ... clarification by Ady TZIDON)

In the sample case given, g=14, b=6 and 2k=8, violating that g<2k.

Be that as it may:

We need to know the probability of the number of girls receiving 5, 6, 7 or 8 tickets.

5 tickets: (14/20)*(13/19)*(12/18)*(11/17)*(10/16)*(6/15)*(5/14)*(4/13)*C(8,5)
6 tickets: (14/20)*(13/19)*(12/18)*(11/17)*(10/16)*(9/15)*(6/14)*(5/13)*C(8,6)
7 tickets: (14/20)*(13/19)*(12/18)*(11/17)*(10/16)*(9/15)*(8/14)*(6/13)*C(8,7)
8 tickets: (14/20)*(13/19)*(12/18)*(11/17)*(10/16)*(9/15)*(8/14)*(7/13)*C(8,8)

The numbers:
for 5:  .317853457172343
for 6:  .357585139318887
for 7:  .163467492260062
for 8:  .023839009287926

sum:     .862745098039218

Corrected as per Steve Herman's comment (included case of 5 girls receiving tickets).

For the derivation of the above, to take the case of the girls' receiving 5 tickets:

For one particular combination of 5-girl ticket distribution (say gbggbgbg, but the explanation goes better with gggggbbb), there's 14/20 prob. that a girl will actually get that first ticket we'd want to go to a girl, and 13/19 for the second thats hoped to go to a girl, and so on down to 10/16 for the fifth. Then there are only 6 out of the remaining 15 students that are boys; then 5 out of 14 and 4 out of 13. That reaches the end of the 8 tickets to be distributed. Note that we've only covered gggggbbb. There are actually C(8,5) cases where 5 girls and 3 boys get tickets, each with identical probability, and so we multiply by that value.

Edited on October 10, 2014, 7:54 am
 Posted by Charlie on 2014-10-09 09:56:20

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