Question 849370: 1) in how many ways can 10 fruits be split into 3 parts consisting of 4,2,4 fruits respectively.
2)in how many can a committee of 5 Doctors be chosen from 9 Doctors.
3) A basket contains 8 Oranges, 3 Mangoes and 9 Apples. if 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1) in how many ways can 10 fruits be split into 3 parts consisting of 4,2,4 fruits respectively.
If the fruits are all the same the answer is "1" way.
If the fruits are all different the answer is 10C4*6C2*4C4 = 210*15*1 = 3150
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2)in how many ways can a committee of 5 Doctors be chosen from 9 Doctors.
Ans: 9C5 = 9C4 = (9*8*7*6)/(1*2*3*4) = 252 ways
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3) A basket contains 8 Oranges, 3 Mangoes and 9 Apples. if 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.
# of ways to get 1 of each:: 8*3*9 = 219 ways
# of randomly chosen groups of 3: 20C3 = (20*19*18)/(1*2*3) = 1140
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Cheers,
Stan H.
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Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
Those answers above are all wrong.
1) in how many ways can 10 fruits be split into 3 parts consisting of 4,2,4 fruits respectively.
Suppose the first part consisting of 4 is labeled part A
Suppose the second part consisting of 2 is labeled part B
Suppose the third part consisting of 4 is labeled part C
If it doesn't matter what fruits are in what part then the answer
would be C(20,4)C(16,2)C(12,4) but since part A and part C are
indistinguishable, we must divide by 2! or 2.
[C(20,4)C(16,2)C(12,4)]/2! = (4845*120*495)/2 = 287793000/2 = 143896500
2)in how many ways can a committee of 5 Doctors be chosen from 9 Doctors.
That's just C(9,5) = 126. (The other tutor set it up right, but worked it
wrong and got twice that.)
3) A basket contains 8 Oranges, 3 Mangoes and 9 Apples. if 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.
We find the probability of the complement event. That's the number of ways to
fail to get 1 of each fruit.
First we find the number of ways to fail to get 1 of each fruit =
The number of ways to get all three the same fruit or 2 of 1 and 1 of another)
P(3 oranges)+P(3 mangos)+P(3 Apples) +
P(2 oranges and 1 mango) + P(2 oranges and 1 apple) + P(2 mangoes and 1 apple) +
P(1 oranges and 2 mango) + P(1 oranges and 2 apples) + P(1 mangoes and 2 apples)
C(8,3)+C(3,3)+C(9,3)+
C(8,2)C(3,1)+C(8,2)C(9,1)+C(3,2)C(9,1)+
C(8,1)C(3,2)+C(8,1)C(9,2)+C(3,1)C(9,2) =
56+1+84 + 84+252+27 + 24+288+108 = 924
The number of ways to select any 3 fruits from the 20 is C(20,3)
P(fail to get 1 of each) = 924/1140 = 77/95 = 0.8105263158
P(succeeding by getting at least 1 of each) = 1-0.8105263158 = 0.1894736842
Edwin
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