Probability of picking correct keys for houses

[kmwest]

Homework Statement

Any help will be appreciated for this homework problem. It's the only one in the set that I haven't finished. :)

"A real estate agent has 8 master keys to open _several_ new homes (underline mine). Only 1 master key will open any given house. if 40% of the homes are usually left unlocked, what is the probability that the real estate agent can get into a specific home if the agent selects 3 master keys at random before leaving the office?"

Homework Equations

Here's what I have so far:

Let A = event that agent can get in a specific home
Let P(U)= probability the given house is unlocked. P(U)=0.4, P(U')=0.6
Let P(K)= probability that one of the three keys the agent selects, opens the desired specific home

The Attempt at a Solution

I'm viewing events U and K as separate, independent events. So I have:

P(A)= P(U)P(K) + P(U)P(K') + P(U')P(K)

Since there are only two options for K, K and K', I can merge the first two terms.

P(A) = P(U) + P(U')P(K)


I am unsure of how to determine P(K). I'm thinking it will be P(K)= (x choose 1)/(8 choose 3). This accounts for the number of possible house choices (x) and number of potential key combinations. So if x=7, that makes P(K)=7/56 and

P(A) = 0.4 + 0.6*0.125 = 0.475

I'm quite unsure about this so please confirm or deny my method... thanks.

 

[Dick]

You can probably assume that one of the eight keys will open the house he chose. He took three of them. What are the odds he picked the right one? I think your answer for P(K) is wrong. It's simpler than you think.

 

[lanedance]

so p(unlocked)=0.4

and couldn't you just say p(correct key)=3/8...

 

[lanedance]

and then consider them independent as you have...

 

[kmwest]

You can probably assume that one of the eight keys will open the house he chose. He took three of them. What are the odds he picked the right one? I think your answer for P(K) is wrong. It's simpler than you think.

So would P(K)= 3/x , assuming x<= 8? I'll be honest, my first thought was P(K)=3/8, but the wording in the question of "several," and not precisely 8, houses has thrown me off.

 

[phinds]

My old brain no longer does well with formulas, but it seems to me it would be .4 + (3/8)*.6 = 62.5%

Using the same logic, if he brought 7 of his 8 keys, it would be 92.5% and if he brought only one key it would be 48%, both of which seem intuitively to be quite reasonable (that doesn't make them right, it's just a decent sanity check on the method).

Again using the same logic, if the chance of a house being unlocked were zero, this method reduces to n/8 where n=the number of keys brought (<=8) which makes sense.

Again using the same logic, if the chance of a house being unlocked were 100%, this method reduces to (1.0 + (n/8)*0.0) = 100% regardless of the number of keys brought, and again it makes sense.

 

[Dickfore]

In the bunch of 8 keys, only 1 is the correct one. The probability to pick it from the first try is 1/8, whereas for not picking that one is the opposite 7/8. Suppose you did not pick the correct one. Then, the probability of picking it is 1/7 in the second try and not picking it is 6/7. Similarly, the probability of picking it in the third try is 1/6. Using the total probability formula, the probability of picking the correct key in 3 tries is:

$$\frac{1}{8} + \frac{7}{8} \cdot \frac{1}{7} + \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{1}{6} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$$

So, you have an error in the estimate of this probability in your calculation ($0.125 = 1/8$).

 

[phinds]

In the bunch of 8 keys, only 1 is the correct one. The probability to pick it from the first try is 1/8, whereas for not picking that one is the opposite 7/8. Suppose you did not pick the correct one. Then, the probability of picking it is 1/7 in the second try and not picking it is 6/7. Similarly, the probability of picking it in the third try is 1/6. Using the total probability formula, the probability of picking the correct key in 3 tries is:

$$\frac{1}{8} + \frac{7}{8} \cdot \frac{1}{7} + \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{1}{6} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$$

--- DELETED (see below) ---

LATER: OOPS ... I see now that you were not correcting me, you were explaining to the OP why his solution was wrong and mine is right, although it puzzles me why you think all that arithmetic is necessary to see that if you have 3 keys out of 8 and one of the 8 must be right, then you have a 3/8 chance of one of them being right.

 

[Dickfore]

--- DELETED (see below) ---

LATER: OOPS ... I see now that you were not correcting me, you were explaining to the OP why his solution was wrong and mine is right, although it puzzles me why you think all that arithmetic is necessary to see that if you have 3 keys out of 8 and one of the 8 must be right, then you have a 3/8 chance of one of them being right.

I used all that arithmetic because no one gave a conclusive argument why the probability is 3/8, although it is intuitively plausible. The complication arises because we draw from the pile of 8 keys without returning the drawn keys. The multiplication clearly shows how numerators and denominators of successive fractions cancel and can easily be generalized.

 

[kmwest]

Thanks to all. I was hung up on the unspecified number of houses but now I realize that it doesn't matter, assuming that the house you are trying to enter has a key among the 8 keys! On to the next problem set...

 

[phinds]

I used all that arithmetic because no one gave a conclusive argument why the probability is 3/8, although it is intuitively plausible. The complication arises because we draw from the pile of 8 keys without returning the drawn keys. The multiplication clearly shows how numerators and denominators of successive fractions cancel and can easily be generalized.

Very reasonable.