Question 1085166: A lecturer wants to select two students at random from a class of 7 males and 3 females. What is the probability the two students chosen are female?
Select one:
A. 0.09
B. 0.22
C. 0.3
D. 0.67
Found 3 solutions by MathLover1, jim_thompson5910, MathTherapy:
Answer by MathLover1(20850)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
Let
A = number of ways to choose 2 females
B = number of ways to choose 2 people (any gender)
Order will not matter.
There are 3 ways to pick 2 females. Let's consider the group of females {X,Y,Z} where the letters are codenames for their actual names
The three groups we can form are: {X,Y}, {Y,Z}, {X,Z}
Each group in a set of curly braces
Note how order does not matter. Something like {X,Y} is the same as {Y,X}
An alternative viewpoint is to consider that there are 3 ways to not pick a female. The three cases would be {X}, {Y}, and {Z}
If you wish to use a formula, then you can use the nCr combination formula to get
n C r = (n!)/(r!*(n-r)!)
3 C 2 = (3!)/(2!*(3-2)!)
3 C 2 = (3!)/(2!*1!)
3 C 2 = (3*2!)/(2!*1!)
3 C 2 = (3)/(1!)
3 C 2 = (3)/(1)
3 C 2 = 3/1
3 C 2 = 3
So that confirms the previous results
In this case, n = 3 because there are three applicants to choose from and r = 2 because that's the number of selections made
So far, we know that A = 3 because there are three ways to pick two women (from a pool of 3).
---------------------------------------------------------------------------------------------------
Using the same formula, but now with n = 10 (there are 10 people total), we get
n C r = (n!)/(r!*(n-r)!)
10 C 2 = (10!)/(2!*(10-2)!)
10 C 2 = (10!)/(2!*8!)
10 C 2 = (10*9*8!)/(2!*8!)
10 C 2 = (10*9)/(2!)
10 C 2 = (10*9)/(2*1)
10 C 2 = 90/2
10 C 2 = 45
Indicating that there are 45 different unique groups we can form if we do not worry about the gender of who is selected.
This means that B = 45
---------------------------------------------------------------------------------------------------
Now divide the values A and B to get A/B = 3/45 = 1/15
The probability as a fraction is 1/15. Don't forget to fully reduce. This value is exact.
Because your teacher wants the answer in decimal form, you would use a calculator or long division to get 1/15 = 0.06666666666667 which is approximate. That value rounds to 0.067
Because this answer isn't listed, I'm assuming there's a typo somewhere. A guess I have is that choice D should be 0.067 instead of 0.67
Or perhaps choice D should be 0.07 instead of 0.67
In any event, I'm thinking your teacher meant to have the answer be choice D. I would check with the teacher though.
---------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------
Final Answer: choice D) 0.67
Note: Look at the section above for an explanation
Edit: MathLover1's solution makes sense on the surface, but looking at it closer there's a flaw in the logic. The answer (3/10)*(3/10) = 9/100 = 0.09 would only apply if the first applicant chosen is put back into the pool. It doesn't make sense to select the same applicant twice.
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
A lecturer wants to select two students at random from a class of 7 males and 3 females. What is the probability the two students chosen are female?
Select one:
A. 0.09
B. 0.22
C. 0.3
D. 0.67
This is not one of the choices.
|
|
|
