# Math Problem

December 4, 2014 4:26 AM Subscribe

Please help with this probability related math problem.

This should be simple for many people. It seems I've been out of a math class for too long.

"Two teams of rock climbers are each setting up a system of anchors on a cliff. Team A's system consists of three anchor points, where the failure of any one of the three points results in the catastrophic failure of the entire system. Team B's system consists of five anchor points, where failure of any two of the five points results in the catastrophic failture of the entire system. Assuming all anchor points have an equal chance of failing during the duration of the exercise, which team's system is safer, and by how much?"

In case you are curious, the question is related to this: link

This should be simple for many people. It seems I've been out of a math class for too long.

"Two teams of rock climbers are each setting up a system of anchors on a cliff. Team A's system consists of three anchor points, where the failure of any one of the three points results in the catastrophic failure of the entire system. Team B's system consists of five anchor points, where failure of any two of the five points results in the catastrophic failture of the entire system. Assuming all anchor points have an equal chance of failing during the duration of the exercise, which team's system is safer, and by how much?"

In case you are curious, the question is related to this: link

The probability of an anchor point failing is p. A will fail if any of those fail. The probability that none of them fail is just the probability of one of them not failing multipled by itself 3 times.

So P(A)=(1-p)^3

B fails if 2 out of any 5 fail. So it succeeds only if none fail or only one fails

P(B)=(1-p)^5+5*p(1-p)^4

(using the binomial theorem)

Now we need to determine which is larger:

P(B)-P(A)=(1-p)^3[(1-p)^2+5*p(1-p)-1)>=

(because (1-p)>=0)

(1-p)^2+5*p(1-p)-1=

p^2-2p+1+5p-p^2-1=

3p>=0

So P(B)>=P(A).

And P(B)-P(A)=3p(1-p)^3.

posted by Cannon Fodder at 4:47 AM on December 4, 2014

So P(A)=(1-p)^3

B fails if 2 out of any 5 fail. So it succeeds only if none fail or only one fails

P(B)=(1-p)^5+5*p(1-p)^4

(using the binomial theorem)

Now we need to determine which is larger:

P(B)-P(A)=(1-p)^3[(1-p)^2+5*p(1-p)-1)>=

(because (1-p)>=0)

(1-p)^2+5*p(1-p)-1=

p^2-2p+1+5p-p^2-1=

3p>=0

So P(B)>=P(A).

And P(B)-P(A)=3p(1-p)^3.

posted by Cannon Fodder at 4:47 AM on December 4, 2014

I think it depends on the probability that an anchor point fails. Each of the outcomes EndsOfInvention details are not equally likely--certainly a single anchor is more likely to fail than all five.

posted by hoyland at 4:48 AM on December 4, 2014

posted by hoyland at 4:48 AM on December 4, 2014

The answer to this question really depends on the probability of a single anchor point's failure.

Let p be the probability that a single anchor point fails.

Then in Team A's case:

The system will only not fail if all anchor points hold. The probability of a single anchor point holding is (1-p). Since we are assuming that the chance of one anchor point holding is independent of the others holding, the probability that they all hold is (1-p)^3

Therefore the chance of the system holding is also (1-p)^3

In Team B's case:

The system will not fail (or will hold) if all anchor points hold or if only one anchor point fails. The chance of all anchor points holding is similar to the case above and is (1-p)^5. The chance of one anchor point failing and all others holding is 5_C_1*(1-p)^4*p (since we need to choose 1 anchor point to fail out of 5, then we have (1-p)^4 for the chance of 4 anchor points holding and p for the 1 anchor point that fails). Since these two possibilities for Team B's system holding are mutually exclusive, the probability overall the Team B's system holds is (1-p)^5 + 5*(1-p)^4*p.

Are there situations in which Team A and Team B's systems are equally safe? We can set up a quadratic equation and solve for p:

(1-p)^3 = (1-p)^5 + 5(1-p)^4*p

Canceling (1-p)^3 on both sides we get

1 = (1-p)^2 + 5(1-p)*p

Solving for p we get p = 0.609 -- this implies that when the probability of failure of a single anchor point is 0.609 the two systems are equally safe. When p > 0.609, Team A's system is safer.

On preview, EndsofInvention's answer assumes that all those outcomes have the same probability (which is not the case).

posted by peacheater at 4:50 AM on December 4, 2014

Let p be the probability that a single anchor point fails.

Then in Team A's case:

The system will only not fail if all anchor points hold. The probability of a single anchor point holding is (1-p). Since we are assuming that the chance of one anchor point holding is independent of the others holding, the probability that they all hold is (1-p)^3

Therefore the chance of the system holding is also (1-p)^3

In Team B's case:

The system will not fail (or will hold) if all anchor points hold or if only one anchor point fails. The chance of all anchor points holding is similar to the case above and is (1-p)^5. The chance of one anchor point failing and all others holding is 5_C_1*(1-p)^4*p (since we need to choose 1 anchor point to fail out of 5, then we have (1-p)^4 for the chance of 4 anchor points holding and p for the 1 anchor point that fails). Since these two possibilities for Team B's system holding are mutually exclusive, the probability overall the Team B's system holds is (1-p)^5 + 5*(1-p)^4*p.

Are there situations in which Team A and Team B's systems are equally safe? We can set up a quadratic equation and solve for p:

(1-p)^3 = (1-p)^5 + 5(1-p)^4*p

Canceling (1-p)^3 on both sides we get

1 = (1-p)^2 + 5(1-p)*p

Solving for p we get p = 0.609 -- this implies that when the probability of failure of a single anchor point is 0.609 the two systems are equally safe. When p > 0.609, Team A's system is safer.

On preview, EndsofInvention's answer assumes that all those outcomes have the same probability (which is not the case).

posted by peacheater at 4:50 AM on December 4, 2014

Using Cannon Fodder's p(X) = probability of team X's success notation:

p(B) - p(A) = (1-p)^5 + 5p(1-p)^4 - (1-p)^3

= (1-p)^3((1-p)^2 + 5p(1-p) - 1)

= (1-p)^3(1 - 2p + p^2 + 5p - 5p^2 - 1)

= (1-p)^3(3p - 4p^2)

This is greater than 0 when 3p - 4p^2 > 0, i.e. when 3/4 > p. (I think team B is safer when p < 3/4.)

In other words, either peacheater or I goofed because we solved the same quadratic.

posted by hoyland at 4:56 AM on December 4, 2014 [4 favorites]

p(B) - p(A) = (1-p)^5 + 5p(1-p)^4 - (1-p)^3

= (1-p)^3((1-p)^2 + 5p(1-p) - 1)

= (1-p)^3(1 - 2p + p^2 + 5p - 5p^2 - 1)

= (1-p)^3(3p - 4p^2)

This is greater than 0 when 3p - 4p^2 > 0, i.e. when 3/4 > p. (I think team B is safer when p < 3/4.)

In other words, either peacheater or I goofed because we solved the same quadratic.

posted by hoyland at 4:56 AM on December 4, 2014 [4 favorites]

Cannon Fodder, while I agree with the first part of your answer, you've made a couple mistakes in the algebra here:

In the first step you've said that (1-p)^3[(1-p)^2+5*p(1-p)-1)>=(1-p)^2+5*p(1-p)-1 because (1-p)>=0, which does not follow. (1-p) is >=0, but in fact the LHS of your equation is <>

Then (1-p)^2+5*p(1-p)-1=p^2-2p+1+5p-p^2-1 is not correct (the RHS should be p^2-2p+1+5p-5p^2-1). This means that the p^2 terms do not cancel.

posted by peacheater at 4:58 AM on December 4, 2014

*P(B)-P(A)=(1-p)^3[(1-p)^2+5*p(1-p)-1)>=*

(because (1-p)>=0)

(1-p)^2+5*p(1-p)-1=

p^2-2p+1+5p-p^2-1=

3p>=0

So P(B)>=P(A).(because (1-p)>=0)

(1-p)^2+5*p(1-p)-1=

p^2-2p+1+5p-p^2-1=

3p>=0

So P(B)>=P(A).

In the first step you've said that (1-p)^3[(1-p)^2+5*p(1-p)-1)>=(1-p)^2+5*p(1-p)-1 because (1-p)>=0, which does not follow. (1-p) is >=0, but in fact the LHS of your equation is <>

Then (1-p)^2+5*p(1-p)-1=p^2-2p+1+5p-p^2-1 is not correct (the RHS should be p^2-2p+1+5p-5p^2-1). This means that the p^2 terms do not cancel.

posted by peacheater at 4:58 AM on December 4, 2014

oops, I lost the 5 on my -p^2 earlier, so we actually have

p^2-2p+1+5p-5p^2-1=3p-4p^2=p(3-4p)

So p=3/4. I agree with Hoyland.

Peacheater, I don't think theres any problem with the inequality step, but I lost that 5 making my conclusion wrong.

posted by Cannon Fodder at 5:00 AM on December 4, 2014

p^2-2p+1+5p-5p^2-1=3p-4p^2=p(3-4p)

So p=3/4. I agree with Hoyland.

Peacheater, I don't think theres any problem with the inequality step, but I lost that 5 making my conclusion wrong.

posted by Cannon Fodder at 5:00 AM on December 4, 2014

I agree with Hoyland too, I goofed when solving the quadratic. Team B is safer when p < 3/4.

posted by peacheater at 5:01 AM on December 4, 2014

posted by peacheater at 5:01 AM on December 4, 2014

oh wait, no it should be less than or equal to. Jesus my brain today.... It';s my fault for trying to do this in a browser rather than with pen and paper.

Hoyland's answer is correct.

posted by Cannon Fodder at 5:03 AM on December 4, 2014

Hoyland's answer is correct.

posted by Cannon Fodder at 5:03 AM on December 4, 2014

This thread is closed to new comments.

all OK = 11 fail, 2 OKs = 32 fail, 1 OK = 3

all fail = 1

So:

8 possibilities, 7

bad, 1good7/8 chance of bad outcome

The combinations of OK/fail for B's system are:

all OK = 11 fail, 4 OK = 5

2 fail, 3 OK = 103 fail, 2 OK = 10

4 fail, 1 OK = 5

all fail = 1

So:

32 possibilities, 26

bad, 6good26/32, or 13/16 chance of bad outcome

A's result of 7/8 is 14/16, so A has the worse chance of failure vs B with 13/16.

posted by EndsOfInvention at 4:43 AM on December 4, 2014