Help with probability?
schools drinks machine is playing up. amy selects water from the machine the probability she gets water is 0.6 she uses the machine 3 times selecting water each time What the probability that a) she gets exact 2 bottles of water b) she gets at least 1 bottle of water. I need to learn how to work this out to help with my daughters homework, so not just the answers required. thanks all
Answers:
a) The probability of getting 'exactly' two bottles. The probablility of getting an exact output say three bottles A,B,C where A=water B=water C=other would be .6 times .6 times .4 (the prob of not water) but exactly two bottles of water could happen three different ways, AB, AC, or BC so the result would need to multiplied by three. So (.6 times .6 times .4) times 3 = 0.432 probability..
b) easier. since the only way she would not get at least 1 bottle of water would be for all three bottles to be other the probability would be one minus the probability that all the bottles were NOT water. So 1 minus (.4 times .4 times .4) = 1 minus .064 which equals a 0.936 probability..
Not very probable.
A) For amy to get exactly 2 bottles of water, she needs to
(NB from now on i will call getting a bottle of water 'succeeding' and not getting one 'failing'
a) Succeed, succeed then fail
b) Succeed, fail, then succeed
c) Fail, succeed then succeed.
The probability of (a) happening is: 0.6 x 0.6 x 0.4 = 0.144
The probability of (b) happening is the same (0.6 x 0.4 x 0.6 = 0.144)
The probability of (c) happening is the same again.
So, the overall probability of getting 2 bottles of water is a + b + c =
0.144 + 0.144 + 0.144 = 0.432 = 43.2%
B) The easiest way to look at this is:
the probability of getting at least one bottle = 1 - the probability of getting none.
This is because Amy must either get zero bottles or at least one bottle, so if you add the two together you must get 1, or 100%.
So, for Amy to get zero bottles, she must fail three times. The probability of this happening is:
0.4 x 0.4 x 0.4 = 0.064 = 6.4%
So the probability of getting at least one is:
1 - 0.064 = 0.936
or 100% - 6.4% = 93.6%
Go back to whoever set your daughters homework and get them to apologise profusely..NOW!
The Key is to use the nCr button on your calculator. The rest you can work out for yourself.
There are three different combinations that gives 2 water
w=water,o=other
wow,wwo,oww
The probability of each of these combinations is
0.6*0.4*0.6=0.144
They are all equally probable so multiply by three
3*0.144=0.432=43.2%
At least one bottle of water is the same as 1-P(no bottle of water)
1-0.4*0.4*0.4=0.936=93.6%
Probability of 'water' is 0.6, 'not water' is 0.4. There are three ways she can get 2 waters ( I will use 'W' for water and 'NW' for not water)
Three ways are: (NW, W, W), (W, NW, W) or (W, W, NW).
Probability of each of these is (0.6 x 0.6 x 0.4), so overall probability is 3 x (0.6 x 0.6 x 0.4) which is 0.432.
I think.
Same should apply to at least 1 bottle of water,
Number of ways is exactly 1+exactly 2+exactly 3
Prob exactly 1 is 3 x (0.6 x 0.4 x 0.4) = 0.288
Prob exactly 3 is just (0.6 x 0.6 x 0.6) = 0.216
add em all up and prob is 0,936, which should be the same as
[1-prob of none at all (0.4 x 0.4 x 0.4) ]
which it is.
Phew! Not sure if I did that the easy way, but I think it's right.
To get 1 bottle of water there is a 60% probability as each time Amy presses a button there is a 60% probabiliy of getting a bottle of water.
To get two botles of water you have a 60% chance after the original 60% (I am getting a bit confused now) so if we take 100. 60% of which is 60. 60% of 60 = 36. so 36%?
By the same logic - probability of getting 3 bottles 21.6%?
That is my stab at it!
I am assuming that there are infinitely many bottles in the vending machine, because as the event occurs, the probability changes.
The probability of Amy not getting a bottle after one press is 1-.6=.4.
If she pressed it thrice, then the probability of getting two bottle is .6*.6*.4 because you multiply together the probabilities of an event happening. So for a) the answer is .144.
The probability that she get only one bottle is .6*.4*.4.
So the answer to b) is .096
a) the possible combinations are:
YYN
YNY
NYY
N = no bottle.
If you work out the prob of each of these and add them up, you get your answer
So: 0.6x0.6x0.4 = 0.144
Each scenario above has the same prob, so 0.144 x 3 = 0.432
b) This is the same as saying she cannot get 0 bottles. If you work out the prob of getting 0 bottles and subtract from 1, you get the answer (I think)
Disclaimer: it was a long time since I was at school.
Good luck
also :D
a) you want exactly two bottles so the probability for each hand is 6/10=3/5..
in three turns it is :
3/5 x 3/5 x 2/5 = 18/125, and there are 3 ways you can have it:
xx-, -xx, x-x.. so it is:3 x 18/125 = 54/125
b) it is simpler to think this other way round.. let's say the prob you will have water in all turns is 1. so 1- 8/125 (prob of having none): 117/125.
a) Combinations resulting in 2 bottles -
1st bottle (0.6) x 2nd bottle (0.6) x 3rd non bottle (0.4) = 0.6x0.6x0.4 = 0.144
Same percentage for following combinations
1st bottle (0.6) x 2nd no bottle (0.4) x 3rd bottle (0.6) = 0.6x0.4x0.6 = 0.144
1st no bottle (0.4) x 2nd bottle (0.6) x 3rd bottle (0.6) = 0.4x0.6x0.6 = 0.144
Total probablility = 0.144+0.144+0.144 = 0.432
b) Probability of at least 1
Only one combination results in less than 1 bottle
1st no bottle (0.4) x 2nd no bottle (0.4) x 3rd no bottle (0.4) = 0.4x0.4x0.4 = 0.064. Hence the probability of no bottles = 0.064, and the probability of at least 1 bottle = 1 - 0.064 = 0.936
In each trial, you have two possible outcomes: either Amy gets a bottle of water or she doesn't. Let .6 = the prob she gets the water, and .4 = the prob she doesn't.
You have 8 possible outcomes:
To get water 3 times: .6*.6*.6 = .216
To get 2 bottles:
.6*.6*.4
+ .6*.4*.6
+ .4*.6*.6 = .432
To get 1 bottle:
.4*.4*.6
+ .4*.6*.4
+ .6*.4*.4 = .288
To get 0 bottles: .4*.4*.4 = .064
Add up all of the permutations, and you should get 1.000! From here, you can find the prob for any number of bottles (exactly two, at least one, whatever).
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Answers:
a) The probability of getting 'exactly' two bottles. The probablility of getting an exact output say three bottles A,B,C where A=water B=water C=other would be .6 times .6 times .4 (the prob of not water) but exactly two bottles of water could happen three different ways, AB, AC, or BC so the result would need to multiplied by three. So (.6 times .6 times .4) times 3 = 0.432 probability..
b) easier. since the only way she would not get at least 1 bottle of water would be for all three bottles to be other the probability would be one minus the probability that all the bottles were NOT water. So 1 minus (.4 times .4 times .4) = 1 minus .064 which equals a 0.936 probability..
Not very probable.
A) For amy to get exactly 2 bottles of water, she needs to
(NB from now on i will call getting a bottle of water 'succeeding' and not getting one 'failing'
a) Succeed, succeed then fail
b) Succeed, fail, then succeed
c) Fail, succeed then succeed.
The probability of (a) happening is: 0.6 x 0.6 x 0.4 = 0.144
The probability of (b) happening is the same (0.6 x 0.4 x 0.6 = 0.144)
The probability of (c) happening is the same again.
So, the overall probability of getting 2 bottles of water is a + b + c =
0.144 + 0.144 + 0.144 = 0.432 = 43.2%
B) The easiest way to look at this is:
the probability of getting at least one bottle = 1 - the probability of getting none.
This is because Amy must either get zero bottles or at least one bottle, so if you add the two together you must get 1, or 100%.
So, for Amy to get zero bottles, she must fail three times. The probability of this happening is:
0.4 x 0.4 x 0.4 = 0.064 = 6.4%
So the probability of getting at least one is:
1 - 0.064 = 0.936
or 100% - 6.4% = 93.6%
Go back to whoever set your daughters homework and get them to apologise profusely..NOW!
The Key is to use the nCr button on your calculator. The rest you can work out for yourself.
There are three different combinations that gives 2 water
w=water,o=other
wow,wwo,oww
The probability of each of these combinations is
0.6*0.4*0.6=0.144
They are all equally probable so multiply by three
3*0.144=0.432=43.2%
At least one bottle of water is the same as 1-P(no bottle of water)
1-0.4*0.4*0.4=0.936=93.6%
Probability of 'water' is 0.6, 'not water' is 0.4. There are three ways she can get 2 waters ( I will use 'W' for water and 'NW' for not water)
Three ways are: (NW, W, W), (W, NW, W) or (W, W, NW).
Probability of each of these is (0.6 x 0.6 x 0.4), so overall probability is 3 x (0.6 x 0.6 x 0.4) which is 0.432.
I think.
Same should apply to at least 1 bottle of water,
Number of ways is exactly 1+exactly 2+exactly 3
Prob exactly 1 is 3 x (0.6 x 0.4 x 0.4) = 0.288
Prob exactly 3 is just (0.6 x 0.6 x 0.6) = 0.216
add em all up and prob is 0,936, which should be the same as
[1-prob of none at all (0.4 x 0.4 x 0.4) ]
which it is.
Phew! Not sure if I did that the easy way, but I think it's right.
To get 1 bottle of water there is a 60% probability as each time Amy presses a button there is a 60% probabiliy of getting a bottle of water.
To get two botles of water you have a 60% chance after the original 60% (I am getting a bit confused now) so if we take 100. 60% of which is 60. 60% of 60 = 36. so 36%?
By the same logic - probability of getting 3 bottles 21.6%?
That is my stab at it!
I am assuming that there are infinitely many bottles in the vending machine, because as the event occurs, the probability changes.
The probability of Amy not getting a bottle after one press is 1-.6=.4.
If she pressed it thrice, then the probability of getting two bottle is .6*.6*.4 because you multiply together the probabilities of an event happening. So for a) the answer is .144.
The probability that she get only one bottle is .6*.4*.4.
So the answer to b) is .096
a) the possible combinations are:
YYN
YNY
NYY
N = no bottle.
If you work out the prob of each of these and add them up, you get your answer
So: 0.6x0.6x0.4 = 0.144
Each scenario above has the same prob, so 0.144 x 3 = 0.432
b) This is the same as saying she cannot get 0 bottles. If you work out the prob of getting 0 bottles and subtract from 1, you get the answer (I think)
Disclaimer: it was a long time since I was at school.
Good luck
also :D
a) you want exactly two bottles so the probability for each hand is 6/10=3/5..
in three turns it is :
3/5 x 3/5 x 2/5 = 18/125, and there are 3 ways you can have it:
xx-, -xx, x-x.. so it is:3 x 18/125 = 54/125
b) it is simpler to think this other way round.. let's say the prob you will have water in all turns is 1. so 1- 8/125 (prob of having none): 117/125.
a) Combinations resulting in 2 bottles -
1st bottle (0.6) x 2nd bottle (0.6) x 3rd non bottle (0.4) = 0.6x0.6x0.4 = 0.144
Same percentage for following combinations
1st bottle (0.6) x 2nd no bottle (0.4) x 3rd bottle (0.6) = 0.6x0.4x0.6 = 0.144
1st no bottle (0.4) x 2nd bottle (0.6) x 3rd bottle (0.6) = 0.4x0.6x0.6 = 0.144
Total probablility = 0.144+0.144+0.144 = 0.432
b) Probability of at least 1
Only one combination results in less than 1 bottle
1st no bottle (0.4) x 2nd no bottle (0.4) x 3rd no bottle (0.4) = 0.4x0.4x0.4 = 0.064. Hence the probability of no bottles = 0.064, and the probability of at least 1 bottle = 1 - 0.064 = 0.936
In each trial, you have two possible outcomes: either Amy gets a bottle of water or she doesn't. Let .6 = the prob she gets the water, and .4 = the prob she doesn't.
You have 8 possible outcomes:
To get water 3 times: .6*.6*.6 = .216
To get 2 bottles:
.6*.6*.4
+ .6*.4*.6
+ .4*.6*.6 = .432
To get 1 bottle:
.4*.4*.6
+ .4*.6*.4
+ .6*.4*.4 = .288
To get 0 bottles: .4*.4*.4 = .064
Add up all of the permutations, and you should get 1.000! From here, you can find the prob for any number of bottles (exactly two, at least one, whatever).
The answers post by the user, for information only, UKQnA.com does not guarantee the right.