Question

Rayne Wooney

A probability question have been annoying me all week, I just can't get my head around it.

"A fair dice is thrown 3 times, find the probability that it will land on at least one six."

Any ideas?

MathsManiac

The question wording is not fantastic. I assume it means this:

"A fair die is thrown 3 times. Find the probability that it will land on six at least once."

In cases like this, it's often easier to consider the complement of the event you're looking for.

So, can you see how you might calculate the probability that it does not land on six at all?

Rayne Wooney

Yeah sorry, I had to translate it from Irish. That's the right way to word it.

For the dice not to land on 6 the three times it's

5/6 x 5/6 x 5/6 = 125/216

But I can't see how to do it for "at least one 6"

brianwalshcork

I'll have a guess at:

1-5/6*5/6*5/6 = 81/216

?

Rayne Wooney

brianwalshcork wrote: »

I'll have a guess at:

1-5/6*5/6*5/6 = 81/216

?

That's what I tried as well but it's the wrong answer

Jumblon

3 throws means 6^3 = 216 possibilities.

? of those have 1 six. ?? have 2 sixes. ??? of them have 3 sixes.

Chance of getting at least one six = (? + ?? + ???)/216.

I think.

MathsManiac

Rayne Wooney wrote: »

That's what I tried as well but it's the wrong answer

It's correct apart from a subtraction error: 216 - 125 = 91, not 81, so the answer is 91/216.

Jumblon's method will give the same answer, as follows:
Of the 216 possible combinations, 1 has three sixes, 15 have exactly two sixes, and 75 have exactly 1 six.

(Try to figure out where the 15 and 75 come from, if you like!)

Jumblon

MathsManiac wrote: »

(Try to figure out where the 15 and 75 come from, if you like!)

Oh, that's easy. Just write out all the possibilities and count them up.

Jumblon

So what's the chance of getting 4 sixes?

Do the equations become complex for n>3?

Rayne Wooney

MathsManiac wrote: »

It's correct apart from a subtraction error: 216 - 125 = 91, not 81, so the answer is 91/216.

Jumblon's method will give the same answer, as follows:
Of the 216 possible combinations, 1 has three sixes, 15 have exactly two sixes, and 75 have exactly 1 six.

(Try to figure out where the 15 and 75 come from, if you like!)

Yeah 91/216 was the answer I got but I wasn't sure if it was the correct one, thanks for the help.

Just another one that I've been trying today..

The probability of an archer hitting a target is 0.2. If the archer fires two arrows, what is the probability that only one will hit the target.

Is it something like

1- 2/10 x 8/10 ?

Yakuza

Rayne Wooney wrote: »

Yeah 91/216 was the answer I got but I wasn't sure if it was the correct one, thanks for the help.

Just another one that I've been trying today..

The probability of an archer hitting a target is 0.2. If the archer fires two arrows, what is the probability that only one will hit the target.

Is it something like

1- 2/10 x 8/10 ?

As MM mentioned above, these problems are sometimes easier solved by finding the probability of the opposite of what you're looking for ("the complement of the event") and subtracting it from 1.

In the case of the new problem, he has a probability of hitting a target of 2/10, or a probability of missing it of 8/10 (we'll ignore the chances of a second arrow hitting the first one, and not going into the target )

In two attempts, what's the probability of missing both?
In two attempts, what's the probability of hitting twice?

Working those out, and taking from 1 will give you your answer.

Your approach (calculating it directly) is along the right lines, but you need to take into account that there are two shots - to hit the target exactly once either of the events could happen - Hit/Miss or Miss/Hit. Each of these has probabilty 2/10 * 8/10. Can you see what you need to do to correct your attempt?
By calculating it directly you don't need to take the answer from 1

Jumblon wrote: »

So what's the chance of getting 4 sixes?



Do the equations become complex for n>3?

Not really, if you google "Bernoulli Trials", you'll get lots of hits with examples.

Rayne Wooney

Yakuza wrote: »

we'll ignore the chances of a second arrow hitting the first one, and not going into the target

I think my head will explode if we go near that

Yakuza wrote: »

In two attempts, what's the probability of missing both?
In two attempts, what's the probability of hitting twice?

Working those out, and taking from 1 will give you your answer.

Your approach (calculating it directly) is along the right lines, but you need to take into account that there are two shots - to hit the target exactly once either of the events could happen - Hit/Miss or Miss/Hit. Each of these has probabilty 2/10 * 8/10. Can you see what you need to do to correct your attempt?
By calculating it directly you don't need to take the answer from 1

Missing both: 0.64?
Hitting Twice: 0.04?

So the probability of 1 of the 2 arrows hitting the target is

(2/10 x 8/10) + (2/10 x 8/10)

1 - 0.32 = 0.68 ?

Yakuza

Rayne Wooney wrote: »

I think my head will explode if we go near that
Missing both: 0.64?
Hitting Twice: 0.04?

So the probability of 1 of the 2 arrows hitting the target is

(2/10 x 8/10) + (2/10 x 8/10)

1 - 0.32 = 0.68 ?

You're almost there.

Yes, the probability of missing twice is 0.64 and the probability of hitting twice is 0.04, so the probability of hitting once is 1- (0.64+0.04) = .32.

Working it out directly - the probabilty of hitting once and missing once is 2/10 * 8/10 = 0.16 *but* this can happen two ways (hit/miss or miss/hit) so the probabilty has to be doubled to take this into account, so the answer is 2 * 0.16 or 0.32, as above.

When there are only two events to consider, it's fairly easy to write out the whole sample space (the fancy shmancy probabilty term for "the entire set of events that could happen in this problem") and work out the relevant probabilities:

Hit / Hit = 0.04
Miss / Miss = 0.64
Miss / Hit = 0.16
Hit / Miss = 0.16
(the above sums to 1, the sum of probabilites of events where there is exactly one hit is 0.32)

but when there are many events (or trials, to use the vocabulary in the Bernoulli Trials link in my above post) and you're asked something like "what is the probability of at less than 8 hits in 10 shots", then writing out the sample space and summing up what you're looking for could get quite tedious, so checking out samples of problems involving Bernoulli Trials and the Binomial Distribution should give you some more pointers.

MathsManiac

Jumblon wrote: »

So what's the chance of getting 4 sixes?

0

If you roll a fair die 3 times, you'd be doing well now to get a six four times, wouldn't ya!

Jumblon

Ah, that's just a hand waiving argument. I want some proper equations.

Jumblon

I got it now. it's 0 because the factorial for negative integers is infinite.