Every year, 6 Irish families, represented by 4 members each, meet up and rent 3 minivans of 8 seats each to go to a big festival. In order to mix things up a little, each individual selects a random car, i.e. family members don’t deliberately try to be in the same car, although they might by chance. The cars go at different speeds, which means they arrive at the festival in a fixed order.
(a) In the first year, what is the probability that in the first car there is exactly one member of the O Chonaill family, given that the second car has exactly two members of that family?
Let Cin = { event that car i has n members of the O Chonaill family }
Let S = { sample space }
|S| = ( 24 C 8,8,8 ) = ( 24 C 8 )*( 16 C 8 )*( 8 C 8 )
P(C22) = probability that 2nd car has 2 O Chonaill
Number of outcomes (where 2 O Chonaill in 2nd car):
1 O Chonaill is in first car, 1 is in third
= [(2 C 1)*(22 C 7)] * [(3 C 2)*(13 C 6)] * (8 C 8)
2 O Chonaill are in first car
= [(20 C 6)] * [(4 C 2)*(13 C 6)] * (8 C 8)
2 O Chonaill are in third car
= [(20 C 6)] * [(4 C 2)*(13 C 6)] * (8 C 8)
Thus the total number of outcomes is the sum of these outcomes:
[(2 C 1)*(22 C 7)] * [(3 C 2)*(13 C 6)] * (8 C 8) + 2*[(20 C 6)] * [(4 C 2)*(13 C 6)] * (8 C 8)
P(C22C11)
number of outcomes:
where 1 is in first car, 2 are in second car, and 1 is in third card:
[(2 C 1)*(22 C 7)] * [(3 C 2)*(13 C 6)] * (8 C 8)
Final Answer:
P(C22C11|C22) =
[(2 C 1)*(22 C 7)] * [(3 C 2)*(13 C 6)] * (8 C 8)
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( [(2 C 1)*(22 C 7)] * [(3 C 2)*(13 C 6)] * (8 C 8) + 2*[(20 C 6)] * [(4 C 2)*(13 C 6)] * (8 C 8) )
Which gives a probability of: 0.6875
(That sequence of dashes is a division bar, it appeared too ugly using a forward slash).
(b) In the second year, assume the people in the first car get out one by one. What is the probability that Briain Mac Carthaigh is among the first five to exit the car, given that there is exactly one Mac Carthaigh in that car?
There are 8 people in the car, so there are (8 C 5) total combinations of people that can be the first five to leave the car. For the number of outcomes where Mac Carthaigh is in the first five, we fix Mac Carthaigh and choose from 4 of the remaining 7 people in the car, and thus we arrive at the following probability:
Final Answer:
P(F|1M) = (7 C 4) / (8 C 5) = 5/8
(c) In the third year, 6 people leave the third car one by one. None of them is an O Cheallaigh. What is the probability there is at least one O Cheallaigh in the car? ´
Attempt 1:
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The total outcomes for the remaining pair of people is (18 C 2), since there are 24 people in total, and 6 of them have left the car. Of the 18 remaining people, 4 of them are chea, since we are told that none of the 6 people who exited the car are chea. Therefore, the number of outcomes where we have a pair of people none of which are chea is (14 C 2), because there are 14 remaining people who are not chea. Thus the probability that none are chea is:
(14 C 2)/(18 C 2)
The probability that there is at least 1 is chea is the complement of the probability that none of the 2 remaining are chea:
Final Answer:
1 - (14 C 2)/(18 C 2) = 62/153 or about 0.405
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Attempt 2 using inclusion exclusion:
probability that exactly 1 of the 2 remaining is a chea:
4 options for person A, 14 (non-chea) options for person B:
4*14/(18*17)
4 options for person A, 3 remaining chea for person B:
4*3/(18*17)
Using inclusion exclusion:
probability that person A is an ochea + probability B is an ochea - probability both A and B are ochea
Final Answer:
2*4*14/(18*17) - 4*3/(18*17) or about 0.326
Which of these two attempts is correct, if either?
Am I totally off the loop with any of these questions (hopefully not all! )
Thanks for any and all help!
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