1

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. (a) How small can the number of socks in the drawer be? (b) How small if the number of black socks is even?

Looks to me like this consists of two parts. First find the percentage for pulling a red sock, ie #Red Socks / #total number of socks. Then multiply that by the chance to pull a second red sock, ie (#Red Socks -1) / (#total number of socks -1)

Under these circumstances it appears that the smallest legal number of socks is 4. 3 red 1 black which yields a 75% chance for the first sock and a 2/3rds chance which when combined yields a 50% chance of pulling out 2 red socks.

If we set the # of black socks to 2 that indicates we have to double our number of red socks to 6. But this is wrong since while our first number is still 75% the second chance is 5/7.

I worked it out by trial and got 15 red socks and 6 black socks. I know that there should be way of figuring this out. The first one ends up being 3/4 * 2/3 so the 3's cancel leaving 2/4 and the second one is 5/7 * 7/10 with the 7's canceling leaving 5/10. Suggestions on how to figure it out would be appreciated.

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. (a) How small can the number of socks in the drawer be? (b) How small if the number of black socks is even?

Looks to me like this consists of two parts. First find the percentage for pulling a red sock, ie #Red Socks / #total number of socks. Then multiply that by the chance to pull a second red sock, ie (#Red Socks -1) / (#total number of socks -1)

Under these circumstances it appears that the smallest legal number of socks is 4. 3 red 1 black which yields a 75% chance for the first sock and a 2/3rds chance which when combined yields a 50% chance of pulling out 2 red socks.

If we set the # of black socks to 2 that indicates we have to double our number of red socks to 6. But this is wrong since while our first number is still 75% the second chance is 5/7.

I worked it out by trial and got 15 red socks and 6 black socks. I know that there should be way of figuring this out. The first one ends up being 3/4 * 2/3 so the 3's cancel leaving 2/4 and the second one is 5/7 * 7/10 with the 7's canceling leaving 5/10. Suggestions on how to figure it out would be appreciated.

2

Successive wins

To encourage Elmer's promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion- father or champion-father-champion, according to Elmer's choice. The champion is a better player than Elmer's father. Which series should Elmer choose?

Since Elmer is more likely to best his father than to best the champion, f is larger than c, and 2-f is smaller than 2-c , and so Elmer should choose CFC. For example, for f=0.8 , c=0.4, the chance of winning the prize with FCF is 0.384, that for CFC is 0.512. Thus the importance of winning the middle game outweights the disadvantage of playing the champion twice.

Successive wins

To encourage Elmer's promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion- father or champion-father-champion, according to Elmer's choice. The champion is a better player than Elmer's father. Which series should Elmer choose?

Since Elmer is more likely to best his father than to best the champion, f is larger than c, and 2-f is smaller than 2-c , and so Elmer should choose CFC. For example, for f=0.8 , c=0.4, the chance of winning the prize with FCF is 0.384, that for CFC is 0.512. Thus the importance of winning the middle game outweights the disadvantage of playing the champion twice.

3

The flippant Juror

A 3-man jury has two members each of whom independently has probability p of making the correct decision and third member who flips a coin for each decision (majority rules). A 1-man jury has probability p of making the correct decision. Which jury has the better probability of making the correct decision?

To calculate the probability that the 3-man jury make a correct decision:

1. when the first two jurors agree with each other: p^2 (the thirds opinion does not matter)

2. when the first two are opposing each other: (p(1-p) +(1-p)p)1/2 The 1/2 is because that the coin favors the correct side half of the time.

Adding the two get the total that is p^2+p(1-p)=p, which is the same as the one-man jury.

The flippant Juror

A 3-man jury has two members each of whom independently has probability p of making the correct decision and third member who flips a coin for each decision (majority rules). A 1-man jury has probability p of making the correct decision. Which jury has the better probability of making the correct decision?

To calculate the probability that the 3-man jury make a correct decision:

1. when the first two jurors agree with each other: p^2 (the thirds opinion does not matter)

2. when the first two are opposing each other: (p(1-p) +(1-p)p)1/2 The 1/2 is because that the coin favors the correct side half of the time.

Adding the two get the total that is p^2+p(1-p)=p, which is the same as the one-man jury.

4

On the average, how many times must a die be thrown until one gets a 6?

Trial Probability of success on trial

1 p

2 pq

3 pq^2

The sum of the probability is

p+pq+pq^2+..... = p(1+q+q^2)

=p/(1-q)

=1

The mean number of trial m is,

m=p+2pq+3pq^2+4pq^3+....

use the trick for summing

qm=pq+2pq^2+3pq^3+...

subtracting the second from the first,

m-qm= p+pq+pq^2+...

mp=1, and m=1/p

so m=6

On the average, how many times must a die be thrown until one gets a 6?

Trial Probability of success on trial

1 p

2 pq

3 pq^2

The sum of the probability is

p+pq+pq^2+..... = p(1+q+q^2)

=p/(1-q)

=1

The mean number of trial m is,

m=p+2pq+3pq^2+4pq^3+....

use the trick for summing

qm=pq+2pq^2+3pq^3+...

subtracting the second from the first,

m-qm= p+pq+pq^2+...

mp=1, and m=1/p

so m=6

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