g 3.24 Socks in a drawer. In your sock drawer you have 4 blue, 5 gray, and 3 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing (a) 2 blue socks (b) no gray socks (c) at least 1 black sock (d) a green sock (e) matching socks

Using the hypergeometric distribution, it is found that there is a:

a) 0.0909 = 9.09% probability that you end up with 2 blue socks.

b) 0.3182 = 31.82% probability that you end up with no gray socks.

c) 0.4545 = 45.45% probability that you end up with at least 1 black sock.

d) 0% probability that you end up with a green sock.

e) 0.2879 = 28.79% probability that you end up with matching socks.

Hypergeometric distribution:

The parameters are:

In this problem:

Item a:

The probability is P(X = 2), hence:

0.0909 = 9.09% probability that you end up with 2 blue socks.

Item b:

The probability is P(X = 0), hence:

0.3182 = 31.82% probability that you end up with no gray socks.

Item c:

The probability is:

In which

0.4545 = 45.45% probability that you end up with at least 1 black sock.

Item d:

There are no green socks, hence 0% probability that you end up with a green sock.

Item e:

Hence, for two gray:

Then, for two black:

Then, the probability of matching socks is:

0.2879 = 28.79% probability that you end up with matching socks.

A similar problem is given at link

a) 1/11 (b) 7/22 (c) 5/11 (d) 0 (e) 19/66

Step-by-step explanation:

Given the following :

Number of Blue socks = n(B) = 4

Number of Gray socks = n(G) =5

Number of black socks = n(Bl) = 3

Total number of socks = (4 + 5 + 3) = 12

Probability = ( number of required outcomes / number of total possible outcomes)

Picking 2 socks at random:

A) probability of two blue socks :

Ist pick = p(B) = (4/12) = 1/3

Number of Blue socks left = (4 - 1) =3

Total socks left = 12 - 1 = 11

2nd pick = p(B) = (3/11)

P(2 blue socks) = (1/3 * 3/11) = 3 /33 = 1/11

B) No gray socks :

Number of non - gray socks = (4 + 3) = 7

1st pick = 7/12

After 1st pick non-gray socks left = 6

Total socks left = 11

2nd pick = 6 / 11

P(non-gray) = (7/12 × 6/11) = 42/132 = 7/22

C.) Atleast one black socks = (1 - P(no black))

Number of non-black socks = (4 +5) = 9

1st pick = 9/12 = 3/4

After 1st pick, non-black left = 8, total = 11

2nd pick = 8/11

P(non - black) = (3/4 × 8/11) = 24/44 = 6/11

P(atleast 1 black) = (1 - 6/11) = 5 /11

D.) A green socks

Number of green socks = 0

P(green) = 0

E.) A matching socks :

1) matching black socks :

Ist pick = p(Bl) = (3/12) = 1/4

Number of Black socks left = (3 - 1) =2

Total socks left = 12 - 1 = 11

2nd pick = p(Bl) = (2/11)

P(matching black socks) = (1/4 * 2/11) = 2 /44 = 1/22

11) matching blue socks:

Ist pick = p(B) = (4/12) = 1/3

Number of Blue socks left = (4 - 1) =3

Total socks left = 12 - 1 = 11

2nd pick = p(B) = (3/11)

P(matching blue socks) = (1/3 * 3/11) = 3 /33 = 1/11

111) matching gray socks :

Ist pick = p(B) = (5/12) = 5/12

Number of Blue socks left = (5 - 1) =4

Total socks left = 12 - 1 = 11

2nd pick = p(B) = (4/11)

P(matching gray socks) = (5/12 * 4/11) = 20/132 = 5 /33

Summing the probabilities :

(1/22 + 1/11 + 5/33) = (3 + 6 + 10) / 66 = 19/66

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: