SOLUTION: The names of four students are placed in a small box and two will be selected to represent the school in an oratory contest. Let 1, 2, 3 and 4 denote the students. What is the pro

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Question 1124321: The names of four students are placed in a small box and two will be selected to represent the
school in an oratory contest. Let 1, 2, 3 and 4 denote the students. What is the probability that: a) 3 is selected?
b) 3 or 4 is selected?
c) 3 is not selected?
Found 3 solutions by Edwin McCravy, Theo, ikleyn:
Answer by Edwin McCravy(20066)   (Show Source): You can put this solution on YOUR website!
The names of four students are placed in a small box and two will be selected
to represent the school in an oratory contest. Let #1, #2, #3 and #4 denote the students. What is the probability that:
a) #3 is selected? 
1 student out of 4 students = 1/4

b) #3 or #4 is selected?
2 students out of 4 students = 2/4 which reduces to 1/2

c) #3 is not selected? 
3 students (#1. #2, or #4) out of 4 students = 3/4

Edwin
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!

the number of ways you can get 2 people out of 4 is 6.

those ways are:

1,2
1,3
1,4
2,3
2,4
3,4

3 out of the 6 contain a 3, so the probability that 3 is selected is 3/6 = 1/2.

5 out of the 6 contain a 3 or a 4, so the probability that 3 or 4 is selected is 5/6.

3 out of the 6 do not contain a 3, so the probability that 3 is not selected is 3/6 = 1/2

you can also do this by formula.

p(3) on the first draw is 1/4
p(not 3) on the second draw is 3/3.
p(3 on the first draw and not 3 on the second draw) is therefore 1/4 * 3/3 = 3/12 = 1/4
p(not 3) on the first draw) is 3/4.
p(3 on the second draw) is 1/3
p(not 3 on the first draw and 3 on the second draw) is therefore 3/4 * 1/3 = 3/12 = 1/4

p(3 on the first draw) is 1/4.
p(3 on the second draw) is 1/4
p(3 on the first or second draw) is therefore 1/4 + 1/4 = 2/4 = 1/2.

p(not 3 on the first or second draw) will be 1 - 1/2 = 1/2.

p(3 or 4) will be p(3) + p(4) - p(3 and 4).

p(3) = 1/2
p(4) = 1/2

p(3 on the first draw and 4 on the second draw) = 1/4 * 1/3 = 1/12.
p(4 on the first draw and 3 on the second draw) = 1/4 * 1/3 = 1/12.

p(3 and 4) on draw 1 or draw 2 is therefore 1/12 + 1/12 = 2/12 = 1/6.

p(3 or 4) is equal to p(3) + p(4) - p(3 and 4).
that would be 1/2 + 1/2 - 1/6 = 5/6.

the formulas provide you with the same answer, but it's not all that clear which formulas to use and how to use them unless you're intimately familiar with their use, which is not always the case.

there's another way using combination formulas.

p(3) would be (1c1) * (3c1) / (4c2) = 3 / 6 = 1/2

p(not 3) = 1 - p(3) = 1/2

p(3 or 4) = p(3) + p(4) - p(3 and 4).
p(3) = (1c1) * (3c1) / (4c2) = 1/2
p(4) = (1c1) * (3c1) / (4c2) = 1/2
p(3 and 4) = (1c1) * (1c1) / (4c2) = 1/6
p(3 or 4) = 1/2 + 1/2 - 1/6 = 5/6

ncx is the combination formula for how many ways you can get sets of x elements out of n elements.

this formula is also shown as c(n,x).

p(3) = 1c1 * 3c1 / 4c2 as shown above.

the 1c1 means you have 1 way to get a 3.
the 3c1 means you have 3 ways to get a number that is not a 3.
the 4c2 means you have 6 ways to get any 2 numbers out of 4.

the calculsations for ncx are n! / (x! * (n-x)!)

for 1c1, that becomes 1! / 1! * 0! which is equal to 1 / 1 * 1 = 1/1 = 1
0! is always equal to 1.

the calculations for 3c1 are 3! / (1! * 2!) = (3*2*1) / (1*2*1) = 6/2 = 3.

so, 3 ways i know of.
first is to work it out.
second is probability formulas
third is combination formulas.

as i said before, they're not all that intuitive.
the more you work with them the better.
when the problems are more complex, i usually try to break them down into something simpler and work it out so i can see what's going on.

at any rate, your solutions are:

p(3) = 1/2
p(4) = 1/2
p(3 or 4) = 5/6
p(not 3) = 1/2


why does the p(3 or 4) = p(3) and p(4) - p(3 and 4).
it's because the p(3 and 4) is double counted so one occurrence of it needs to be subtracted.

consider

the possibilities are:

1,2
1,3
1,4
2,3
2,4
3,4

p(3) contains 1,3 and 2,3 and 3,4
p(4) contains 1,4 and 2,4 and 3,4

the 3,4 is being counted twice; in p(3) and in p(4) and so one of those occurrences needs to be subtracted so there is no double counting.







Answer by ikleyn(52932)   (Show Source): You can put this solution on YOUR website!
.
The names of four students are placed in a small box and two will be selected to represent the
school in an oratory contest. Let 1, 2, 3 and 4 denote the students. What is the probability that: a) 3 is selected?
b) 3 or 4 is selected?
c) 3 is not selected?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


a).   What is the probability that  #3  is selected ? 

    In all, there are   =  = 6 ways to form the groups of 2 students from 4 students.


    And there are only 3 groups of two students that contain #3 as a member, because you can add one 
    of the 3 students #1, #2 and/or #4 to the #3 in three ways.


    So, the probability under this question is   = .


b).   What is the probability that  #3  or  #4  is selected ? 

     If NEITHER  #3  NOR   #4  is selected, then there is only one such a group of two students: it is the group comprising of  #1  and  #2.


     The total number of groups of 2 from 4 is 6, as we calculated it in the solution a).


     So the probability under this question is   = .


c).   What is the probability that  #3  is not selected? ? 

     The number of groups of 2 students from 4 that do not contain #3 is   =  = 3.


     The entire number of groups by 2 from 4 is 6, as we calculated it in the solution a).


     So the probability under this question is   = .


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