# statistic as

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#1
in class 12 A there are 40 students
10 students have blue eyes, 3 students have green eyes , 27 students have brown eyes.

in class 12 B there are 50 students.
22students have blue eyes , 2 students have green eyes , 26 students have brown eyes

one person is chose at random from each class
find the probability :

1) that both students have green eyes
P(both blue)

2) That both students have different color eyes
P(different color) = ?

3) P( one brown) =?
0
6 years ago
#2
(Original post by villain20)
in class 12 A there are 40 students
10 students have blue eyes, 3 students have green eyes , 27 students have brown eyes.
in class 12 B there are 50 students.
22students have blue eyes , 2 students have green eyes , 26 students have brown eyes

one person is chose at random from each class
find the probability :

1) that both students have green eyes
P(both blue)

2) That both students have different color eyes
P(different color) = ?

3) P( one brown) =?
While it isn't essential, have you tried drawing a tree diagram where the first event is a person picked from class A and the second event is a person picked from class B?
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#3
(Original post by gdunne42)
While it isn't essential, have you tried drawing a tree diagram where the first event is a person picked from class A and the second event is a person picked from class B?

Yes i did it but it didnt work.
0
6 years ago
#4
(Original post by villain20)
Yes i did it but it didnt work.
Show me, then I can see where you went wrong

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#5

I did this and i dont know what i should do to make point 2 and 3
0
6 years ago
#6
(Original post by villain20)

I did this and i dont know what i should do to make point 2 and 3
Why are there only two possibilities for the second choice?
If you pick someone from class A with blue eyes you could pick someone from class B with blue eyes.
Your tree really should show all three possible outcomes from class B on every branch to enable you to then answer any possible question you are asked.

1. I am assuming you have correctly calculated it is the probability(A=blue) x probability(B=blue)

2. To tackle part 2 you have two choices. You can either calculate the total of all 6 probabilities where the two are different. Alternatively, remembering the probabilities must add to 1 you can calculate the total probabilities of the 3 outcomes where both are the same and subtract it from 1

3. For part 3 you could add together all of the probabilities where the person from class A has brown eyes but the person from class B doesn't to the outcomes where the class A person doesn't have brown eyes and the class B person does.
There are other ways to go about it that are more efficient that you may be able to think of.

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#7

I did it like this and try 2 point and it still wrong
( i did it on other numbers to see if it work because ive got 2 set of question like this and it still doesnt )
0
6 years ago
#8
(Original post by villain20)

I did it like this and try 2 point and it still wrong
( i did it on other numbers to see if it work because ive got 2 set of question like this and it still doesnt )
Don't know what you've done to get the wrong answers for the original question as you haven't shown your calculations or results and don't know what the second question is so can't help.

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