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23 votes
23 votes
An excursion group of four is to be drawn from among 5 boys and 6 girls.

Find :
a. the number of ways of choosing the excursion group if the group :

i. Is to be made up of an equal number of boys and girls.
ii. Is to be either all-boys or all-girls.
III. Has no restriction on its composition.
b. Whatis the probability that a random choice of numbers from the group will result in 3 boys and 1 girl.​

User FRL
by
2.9k points

2 Answers

4 votes
4 votes

Answer:

Because you are recounting people several times. Say for example, let us number those people

{1,2,3,4} for the four girls and {1,2…7} for the boys.

One possible combination by your way is to fix one boy and girl each, say

{1,1} and then choose the remaining people. Let this combination be, for the sake of the example, {1,1,2,2,3}. This is a valid combination.

Now, when you fix another combination of a boy and a girl, say {1,2}, you could get the same combination as before because one the combinations while choosing the three remaining people would be {1,2,3}.

A correct way would be to choose girls and then choose 5− boys for 1≤≤4

So we get ∑4=1(4)(75−)=441

User Sri Sris
by
2.8k points
14 votes
14 votes

Answer:

We need group of 4 out of group of 5 + 6 = 11 people.

Part A

i)

2 boys and 2 girls:

  • 5C2 * 6C2 = (5*4/2) * (6*5/2) = 150 ways

ii)

4 boys or 4 girls:

  • 5C4 + 6C4 = 5 + 15 = 20 ways

iii)

No restrictions:

  • 11C4 = (11*10*9*8)/(4*3*2) = 330 ways

Part B

Combination of 3 boys and 1 girl:

  • 5C3*6C1 = (5*4/2)*6 = 60

Total number of ways is 330 (found above)

Required probability:

  • P(3 boys and 1 girl) = 60/330 ≈ 0.1818 = 18.18%
User Uluk Biy
by
3.3k points