Probability question, help? +steps with it (changed)

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Ashnil has a security code on his mobile phone. Each time he switches his phone on he has to enter the code.
The probability that he enters the code correctly on his first attempt is 0.9.
If he enters the code incorrectly, the probability that he enters the code correctly on his next attempt is 0.7. Each time he turns his phone on, Asnil is allowed 3 attempts to enter the code. Ashnil turns his phone on. Calculate the probability that Ashnil enters the correct code on his first attempt, second attempt or his third attempt.

Use a branching tree diagram.

There are three mutually exclusive events corresponding to a successful log-in.

1. He gets it right the first time. Probability = 0.1

2 He gets it right on the second try. 0.3

3. He gets it right on the third try 0.3

Add them up and get 0.7

ruveyn

First attempt: 90%

"by second attempt": 97%

"exactly on" second attempt: 7%

That is not correct, you cannot simply add probabilities. If instead of 0.7 the probability of getting it right were 0.5, doing as you say you would get 1.1 which has no sense.



The probability on the first attempt is 0.9 as stated.

The probability on the second attempt is the probability of fail the first (0.1) and then get right the second (0.7), that is P(2)=0.1*0.7=0.07.

The probability on the third attempt is the probability of fail the first (0.1) then fail the second (0.3) and then be correct on the third (0.7), that is, P(3)=0.1*0.3*0.7=0.021.

The only way not to log in is to fail three times in a row.

The probability of the is 0.9 x 0.7 x 0.7 which is about .44. 1.0 minus that is .56 which is the probability of success.

More generally if the probability of failure on try 1 is p1 the probability of failure on try 2 is p2 and the probability of failure onf try 3 is p3, the the probability of failure three times in a row is p1 x p2 x p3. Succeeding on either try 1 or try 2 or try 3 is is the same not failing three times straight which is 1 - p1xp2xp3


ruveyn

Wait. That doesn't make sense. Which is probably because your equation is all wrong. If I'm not mistaken, you use multiplication when you must have probability a happen AND probability b happen. In order to succeed, you only need succeed in entering your password once. But in order to fail you must fail all three times.

So, we take the first probability, 10%, and the second and third failing probabilities, 30%. So it's .1 x .3 x .3 = .009 chance of failing, or a 1 - .009 = 99.1% chance of success. Not 56%. That just doesn't seem to work. He would be locked out of his phone almost half the time, which realistically won't happen (at least on a good phone). And the premises are realistic, so either this is one of those counterintuitive probability things, or ruveyn is wrong. I suspect the latter.

Also, I'm pretty sure the original problem never said what the probability of the third password success was. Assuming .7 like everyone else:

Probability of success on first try:

.9

Probability of success on second try:

failure on first try * success on second try = .1*.7 = .07

Probability of success on third try:

failure on first try * failure on second try * success on third try = .1*.3*.7 = .021

Probability of outright failure and phone lockout:

.009 (see above)

Of course, I can't guarantee any of this is correct, cause I tend to make careless mistakes with this kind of thing, but I'm pretty sure I got it right.


Edit: Sorry, NakaCristo, I didn't see that you had already gotten the right answer already. I have a bad habbit of looking at the top post and then the bottom but skipping over the middle with these kinds of threads, because I figure if the bottom guy got it wrong then no one in the middle got it right.

Last edited by physicsnut42 on 14 Oct 2012, 8:16 am, edited 1 time in total.

Is this homework?