In a diving competition each diver has...

[stevenstratmann]

Guide 4 p 93 #8
In a diving competition each diver has a 20% chance of a perfect dive.The first perfect dive of the competition,but no subsequent dives will recieve a perfect score.If Janet is the third diver to dive,what are her chances of recieving a perfect score?

A pretty straight forward solution is .8 x .8 .2 =.128 x 100= 12.8%

I wached a session of 'Thursdays with Ron' and it appeared a similar question was asked-

On his drive to work, Leo listens to one of three radio stations, A, B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station, the probability is .30 that at a any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo will hear a song he likes?

A .027
b .090
c .417
D .657
e. .900

the solution to this was 3/10 + (7/10 * 3/10) + (7/10 * 7/10 * 3/10)

my question is why can you not use this same theory on the initial question and go: 1/5 + (4/5 *1/5) + 1/5 and get the answer you are looking for?

[cpallaka]

Let me try to help you yo understand difference between these two questions. first I will explain question in studay hall.

if you rephrase the question when does Leo listens to the song he likes?

1. Leo turns to A, listens a song and he likes the song
OR
2. Leo turns to A, listens a song, he does not like it and turns to B and he likes the song
OR
3. Leo turns to A, listens a songand he does not like it; turns to B and he does not like the song; turns to C and likes the song
so you calucalte probabalitiy for each scenario and add them as Ron explained in study hall

Rephrase the strategy guide question.
there are 3 divers and what is the probability of 3rd diver get perfect score. To this can happen only when 2 divers before him should not get perfect score and he gets perfect scope. if you compare with other question, this scenario is equivalent to 3rd scenario of other question.

I hope this helps you

[tim]

thanks for the explanation. Steven, please let us know if you need any further help with this one..

[gokyada]

Just adding another approach to music question. Of the three stations, we are looking for at least one station to provide him good music. Consider cases GBG, GGG, BBB, ..... If we know the probability of scenario not providing any good music, then 1-that probability will be the answer for probability of providing good music by any one station. So probability of BBB is 7/10*7/10*7/10 . Probability of getting good music is 1 - (343/1000) = 0.657.

[jnelson0612]

Agreed, thanks for your input.

[david.khoy]

So probability of BBB is 7/10*7/10*7/10 . Probability of getting good music is 1 - (434/1000) = 0.657.

There is a typo here: 7 * 7 * 7 = 343, not 434.

[jnelson0612]

So probability of BBB is 7/10*7/10*7/10 . Probability of getting good music is 1 - (434/1000) = 0.657.

There is a typo here: 7 * 7 * 7 = 343, not 434.

Thanks, you are right! Fixed above.

[adm45]

I believe "stevenstratmann" and I have the same question and "cpallaka" starts to answer it but I don't understand his explanation ("if you compare with other question, this scenario is equivalent to 3rd scenario of other question."). I will use similar example that caused concern. How is the original question about the diving competition different from the Molly question on Word Trans. 4th pg 87? I thought the diving question would be solved like the Molly question was:

(4/5) + (4/5)(4/5)+ (1/5) = 16/25 or 64%.

Thank you!

[jlucero]

I believe "stevenstratmann" and I have the same question and "cpallaka" starts to answer it but I don't understand his explanation ("if you compare with other question, this scenario is equivalent to 3rd scenario of other question."). I will use similar example that caused concern. How is the original question about the diving competition different from the Molly question on Word Trans. 4th pg 87? I thought the diving question would be solved like the Molly question was:

(4/5) + (4/5)(4/5)+ (1/5) = 16/25 or 64%.

Thank you!

First off, your equation there is incorrect. Adding those fractions would give you 1+16/25, which is a sign that your method is wrong. (and coincidentally, is how theoretical physicists and mathematicians find errors in theories- probabilities that are greater than 100%)

Second, cpallaka was equating one part of the radio problem to the entire diving problem. The odds of the third radio station having a good song, requires the first two to be bad. So the probability is BAD and then BAD and then GOOD. Same idea with a perfect score for the third diver.

Which leads to the most important thing you are missing here. When to add vs when to multiply probabilities. You've obviously been reading the strategy guides, so hopefully you can see the AND vs OR probability at play here. You don't want to add probabilities here, because you aren't seeing what is the probability that the first diver messes up OR the second one messes up. It's the probability that one specific thing happens: BAD, BAD, GOOD. So we multiply these odds rather than add them up.

[StephenR114]

I believe the answer is wrong in the book and above to the diver question.

Diver one has a 1/5 chance at a perfect dive (4/5 chance not perfect).

Diver two's chance of a perfect dive is not simply 1/5 because he is dependent on diver one not having a perfect dive (so 4/5 times 1/5 = 4/25).

So Diver threes chance at a perfect dive is 4/5 times 21/25 times 1/5 which equals 84/625. Is there something I'm missing?

[RonPurewal]

when you multiply probabilities, the earlier events are already taken into account. so, it's just (4/5)(4/5)(1/5).

to prove this to yourself, work with simpler examples. like, say you flip a coin two times.
then there are four equally likely possibilities: HH, HT, TH, TT.
it should be straightforward to see that these are equally likely.

to calculate the probability of each one, we just multiply (1/2)(1/2). the second 1/2 is the probability of whatever thing happening on the second toss once the first toss has already happened—which is ALWAYS what you are figuring when you multiply probabilities together.
(according to your logic here, you'd be trying to do something like (1/2)(1/4) for each of these possibilities. it's plain to see that doesn't work.)