A group of 15 youngsters received 5 tickets to a rock happening.
The lucky recipients were determined by a random drawing of names
(no replacement) from a closed bag.
Although there were only 6 girls in the group they jointly received more tickets than the boys.
a. What is the a priori probability of such partition?
b. What would be the result if the drawing was performed with replacements?
c. Define the meaning of the difference b-a.
For the girls to have received more tickets than the boys, exactly 3, 4 or 5 of the tickets would have had to go to girls.
a)
prob(3) = (6*5*4*9*8/(15*14*13*12*11)) * C(5,3) = 240/1001
prob(4) = (6*5*4*3*9/(15*14*13*12*11)) * C(5,4) = 45/1001
prob(5) = (6*5*4*3*2/(15*14*13*12*11)) * C(5,5) = 2/1001
The total of the three is 41/143 ~= .2867132867132867
Calculations by
10 p3=(6*5*4*9*8//(15*14*13*12*11)) * combi(5,3)
20 p4=(6*5*4*3*9//(15*14*13*12*11)) * combi(5,4)
30 p5=(6*5*4*3*2//(15*14*13*12*11)) * combi(5,5)
40 print p3:print p4:print p5:print p3+p4+p5
b) Each probability of a girl receiving a ticket (possibly for a second time for that girl) is 6/15 = 2/5.
prob(3) = (2/5)^3 * (3/5)^2 * c(5,3) = 144/625
prob(4) = (2/5)^4 * (3/5)^1 * c(5,4) = 48/625
prob(5) = (2/5)^5 * (3/5)^0 * c(5,5) = 32/3125
The total is 992/3125 = .31744
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Posted by Charlie
on 2020-05-04 22:54:55 |
a. and b.
