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Probability Problems

MathStudent1999

Junior Member
Joined
Mar 18, 2012
Messages
76
Johnny got Christmas presents from each of his parents, each of his four grandparents, and one aunt and one uncle. He carefully wrote each thank you note, one to each of the donors, thanking each for the specific present they had sent to him. But then he asked his sister to address and mail the thank you notes, and she did not notice which note was intended for which relative. So she just addressed each of the eight envelopes to a different relative at random. What is the probability that exactly 5 of the 8 relatives got the notes that was intended for them? Express you answer as a common fraction.

I just did 1/8 * 1/7 * 1/6 * 1/5 * 1/4 * 2/3 * 1/2 * 1/1 and got 1/20160, but I'm not sure this answer is correct. Can someone check this problem and if this is wrong can someone teach me how to do this question?
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,708
You have expressed that there is only one chance in all the possible combinations. You will have to rethink that.

Let's just start with mom and dad. Forget everyone else.

Address 1st letter to mom. It's right 1/2 or wrong 1/2

If it's right, then dad is 100%. If it's wrong, then dad is 0%

(1/2)*1 + (1/2)*0 = 1/2

This may seem to support your conlusion, however, you may be overlooking something with only two that is exposed by the question, What is the probability of getting exactly one of these right? If you miss one, you must miss a second. Misses come in at least pairs.

Let's just count grandparents. G1, G2, G3, G4. There are 24 ways to addresss these four. Let's assume the letters are in the first order listed. How many do we get for each possible prdering

G1G2G3G4 - 4
G1G2G4G3 - 2
G1G3G2G4 - 2
G1G3G4G2 - 1
G1G4G2G3 - 1
G1G4G3G2 - 2
G2G1G3G4 - 2
G2G1G4G3 - 0
G2G3G1G4 - 1
G2G3G4G1 - 0
G2G4G1G3 - 0
G2G4G3G1 - 1
G3G1G2G4 - 1
G3G1G4G2 - 0
G3G2G1G4 - 2
G3G2G4G1 - 1
G3G4G2G1 - 0
G3G4G1G2 - 0
G4G1G2G3 - 0
G4G1G3G2 - 1
G4G2G1G3 - 1
G4G2G3G1 - 2
G4G3G1G2 - 0
G4G3G2G1 - 0

If I have listed them correctly, we have p(4) = 1/24, p(3) = 0/24, p(2) = 6/24, p(1) = 8/24, p(0) = 9/24. I'm guessing this is not what you might have expected. In particular, 1/24 is the probability of getting them ALL correct.

Give it another go.
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, MathStudent1999!

Johnny got Christmas presents from each of his parents, each of his 4 grandparents, and 1 aunt and 1 uncle.
He carefully wrote thank-you notes, one to each donor, thanking each for the specific present they gave him.
He asked his sister to address and mail the thank-you notes, and she addressed each of them at random.
What is the probability that exactly 5 of the 8 relatives got the notes that was intended for them?
Express your answer as a common fraction.

There are: 8! = 40,320 possible outcomes.

Since exactly five relatives got their respective notes,
. . the other three had a derangement of their notes.

Choose 3 of the 8 relatives: .[SUB]8[/SUB]C[SUB]3[/SUB] = 56 choices.

They can be deranged in 2 ways: .bca, cab

Hence, there are: .56 x 2 = 112 ways.

. . . . . . . . . . . . . . . . . . . . . . 112 . . . . . . 1
Therefore: .P(exactly 5) . = . -------- . = . -----
. . . . . . . . . . . . . . . . . . . . . 40,320 . . . . 360
 

MathStudent1999

Junior Member
Joined
Mar 18, 2012
Messages
76
Hello, MathStudent1999!


There are: 8! = 40,320 possible outcomes.

Since exactly five relatives got their respective notes,
. . the other three had a derangement of their notes.

Choose 3 of the 8 relatives: .[SUB]8[/SUB]C[SUB]3[/SUB] = 56 choices.

They can be deranged in 2 ways: .bca, cab

Hence, there are: .56 x 2 = 112 ways.

. . . . . . . . . . . . . . . . . . . . . . 112 . . . . . . 1
Therefore: .P(exactly 5) . = . -------- . = . -----
. . . . . . . . . . . . . . . . . . . . . 40,320 . . . . 360
Thanks for your help!



I know have another question:Suppose the Yankees play the Mariners in the World Series, in the first team to win four games wins the series.If the probability of the Yankees winning any one game is 6/10, regardless of which staduim the game is played in, what is the probability that the series will take seven games to decide? Express your answer as a decimal to 5 places.

Is the correct answer 0.27684? I got this by dividing 864 by 3125. If this isn't the correct answer, can someone teach me how to do this question?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello again, MathStudent1999!

Suppose the Yankees play the Mariners in the World Series; the first team to win four games wins the series.
If the probability of the Yankees winning any one game is 0.6, regardless of which staduim the game is played in,
what is the probability that the series will take seven games to decide?
Express your answer as a decimal to 5 places.

Is the correct answer 0.27684? . [COLOR=#b000e]Yes![/COLOR]
I got this by dividing 864 by 3125. . Correct!

The question becomes: what is the probability that, in the first six games,
. . the two teams win three games each?

P(3,3) . = . [SUB]6[/SUB]C[SUB]3[/SUB](0.6)[SUP]3[/SUP](0.4)[SUP]3[/SUP] . = . 0.27648


Your answer: . (6!/3!3!)(3/5)[SUP]3[/SUP](2/5)[SUP]3[/SUP] . = . 864/3125 . = . 0.27648
 

MathStudent1999

Junior Member
Joined
Mar 18, 2012
Messages
76
Hello again, MathStudent1999!


The question becomes: what is the probability that, in the first six games,
. . the two teams win three games each?

P(3,3) . = . [SUB]6[/SUB]C[SUB]3[/SUB](0.6)[SUP]3[/SUP](0.4)[SUP]3[/SUP] . = . 0.27648


Your answer: . (6!/3!3!)(3/5)[SUP]3[/SUP](2/5)[SUP]3[/SUP] . = . 864/3125 . = . 0.27648
Thank you very much!!! :D
 
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