Question

Work the following probability sock problem, keeping in mind that this is a problem without replacement. When you take out a sock, it stays out.

Sock Problem:

Your sock drawer is very unorganized. No socks are paired, and they are all just thrown randomly into the drawer. You do know that the drawer has four red socks and four blue socks in it. You want to get some socks to wear in the morning, but you do not want to turn on a light for fear of waking up your family.

• If you draw two, what is the probability of a red pair match?
• If you draw two, what is the probability of a match of any color?
• If you draw three, what is the probability of a match of any color?
• Describe your thinking and process.
• Include your thinking/reasoning on this problem, answers to the questions in fractional form (not decimals or percentages), and the specific thinking path that led you to the answers.

Answer 1

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GIVEN THAT :-

According to the question we have that

drawer containing of 4 red socks and 4 blue socks  

now finding out the following questions asked

FIND OUT :- If you draw two, what is the probability of a red pair match?

NOW solving

P(red pair match)=4C2/8C2

=6/28

=0.214

choose 2 out of 4 socks to get a red pair(4C2)

Total ways are to choose 2 socks out of 8 socks (8C2)

TO FIND :- If you draw two, what is the probability of a match of any color ?

as we done in the above methoed do it similarly

P(any pair match) = 4C2+4C2/8C2

=6+6/28

=0.428

TO FIND :- If you draw three, what is the probability of a match of any color?

FOR three to match we ca take any colour so

Total ways=8C3

there are four cases in selecting

P(All three are red)

P( all three are blue)

P( 1 red,2 blue)

P( 2 red, 1 blue)

so by doing the sum we get

P(All three are red) + P( all three are blue) + P( 1 red,2 blue) +P( 2 red, 1 blue).

=4C3+4C3+4C1*4C2+4C1*4C2

required probability=4+4+4*6+4*6/56

=56/56

=1

as we have in a pair only two socks . we cant chosse three