Probabililty: drawer contains 5 blue, 1 white, 2 yellow ties

A

Aladdin

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Mar 27, 2009
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A drawer contains 5 blue , 1 white and 2 yellow ties.
If we draw 4 ties with replacement .
Calculate the probability of the events:
A: the first tis is white the 2nd is blue the 3rd is yellow the fourth is white.
B:the 4 ties have 4 different colors.
C:The 4 ties have the same color 2 blue and 2 yellow ties.

But what I think there is something wrong in the question ???

b) Reapeat the above questions if the drawing is made without replacements,


what i have done is the following:

With replacement:
p(1/8*5/8*2/8*1/8)=p(A)

Without replacement:
P(A)=0(zero)

With replacement
P(B)=3!(1/8*5/8*2/8*1/8)

Without Replacement:
P(B)=0

With replacement:
P(C)=(5/8*5/8*5/8*5/8)+(1/8*1/8*1/8*1/8)+(2/8*2/8*2/8*2/8)

Without replacement>???


Thank you in advance.
 
Let's do part c without replacement.

The probability there are 2 blue and 2 yellow.

First choice may be blue: 5/8

second choice may be blue: 4/7

third choice may be yellow: 2/6

fourth choice may be yellow: 1/5

But, they can come out in 4!/(2!2!)=6 different ways.

6(5/8)(4/7)(2/6)(1/5)=1/7.

Another way using combination notation:

\(\displaystyle \frac{C(5,2)C(2,2)C(1,0)}{C(8,4)}=\frac{1}{7}\)
 
Thank you . I have a test tomorrow and I must know the answer for these questions :? .
 
Aladdin said:
A drawer contains 5 blue , 1 white and 2 yellow ties.
If we draw 4 ties with replacement .
Calculate the probability of the events:
A: the first tis is white the 2nd is blue the 3rd is yellow the fourth is white. You have only ONE white tie - cannot draw two whites - without replacement
B:the 4 ties have 4 different colors. <<<< You have only 3 colors - you cannot have 4 different colors - with or without replacement
C:The 4 ties have the same color 2 blue and 2 yellow ties.

But what I think there is something wrong in the question ???

b) Reapeat the above questions if the drawing is made without replacements,


what i have done is the following:

With replacement:
p(1/8*5/8*2/8*1/8)=p(A)

Without replacement:
P(A)=0(zero) ..................................correct

With replacement
P(B)=3!(1/8*5/8*2/8*1/8)..................................Incorrect

Without Replacement:
P(B)=0..................................correct

With replacement:
P(C)=(5/8*5/8*5/8*5/8)+(1/8*1/8*1/8*1/8)+(2/8*2/8*2/8*2/8)

Without replacement>???


Thank you in advance.
Click to expand...
Is there any help?????
Click to expand...

Other two (A & B) are really nonsense question - did you even read carefully??
 
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