AnnieOakley:
I need help with the following...

1. A chemist has a 20% and 40% acid solutions. What amount of each solution should be used in order to make 300ml of a 28% acid solution. [I got 180ml and 120ml. Am I correct?]

Absolutely.

2. The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising equal quantities of all 3 samples is made. What is the proportion of milk and water in the mixture?

Karmanaut:

Absolutely.

Thank you but how about the second one? I don't even know how to go about it.

In a primary school, 70% of the boys and 55% of the girls can ride a bicycle. If a boy and a girl are chosen at random, what is the probability that:


(a) both of them can ride a bicycle [Is it 70/100 * 55/100 or 70/200 * 55/199 or neither?]

(b) neither of them can ride a bicycle [Is it 30/100 * 45/100 or 1 - p(both can ride a bicycle)?]

(c) if 150 boys are selected randomly in the school, what is the probability that they can ride a bicycle? [I did 70/100 * 150 = 105. p(both can ride = 105/150 = 7/10). Is it correct?].

Please help @jackpot, laplacian, antoinne, seamione

AnnieOakley:


Thank you but how about the second one? I don't even know how to go about it.

Sorry.
Forgot.
Proportion of milk in 3 samples is 2:3,3:5,5:8
Proportion of water in 3 samples is 1:3,2:5,3:8
Since equal quantities are taken,
Total proportion of milk is 2/3+3/5+5/8=227/120
Total proportion of water is 1/3+2/5+3/8=133/120

Proportion of milk to water: 227:120

AnnieOakley:
In a primary school, 70% of the boys and 55% of the girls can ride a bicycle. If a boy and a girl are chosen at random, what is the probability that:


(a) both of them can ride a bicycle [Is it 70/100 * 55/100 or 70/200 * 55/199 or neither?]

(b) neither of them can ride a bicycle [Is it 30/100 * 45/100 or 1 - p(both can ride a bicycle)?]

(c) if 150 boys are selected randomly in the school, what is the probability that they can ride a bicycle? [I did 70/100 * 150 = 105. p(both can ride = 105/150 = 7/10). Is it correct?].

I don't think the question is complete, as we weren't given a union of sets (considering the two events as sets)
Normally we should have:
P(A∪B)=P(A)+P(B)−P(A∩B).
Where P(A) =70/100 = 7/10 and P(B) =55/100= 5.5/10

If there was a third set containing the percentage of students that can do both then the question is solvable.

I could be wrong though.
Will think through it later.

Karmanaut:

Sorry.
Forgot.
Proportion of milk in 3 samples is 2:3,3:5,5:8
Proportion of water in 3 samples is 1:3,2:5,3:8
Since equal quantities are taken,
Total proportion of milk is 2/3+3/5+5/8=227/120
Total proportion of water is 1/3+2/5+3/8=133/120

Proportion of milk to water: 227:120

Thanks but why does water has a higher proportion value than water?

Karmanaut:

I don't think the question is complete, as we weren't given a union of sets (considering the two events as sets)
Normally we should have:
P(A∪B)=P(A)+P(B)−P(A∩B).
Where P(A) =70/100 = 7/10 and P(B) =55/100= 5.5/10

If there was a third set containing the percentage of students that can do both then the question is solvable.

I could be wrong though.
Will think through it later.

The question is complete.

AnnieOakley:


Thanks but why does water has a higher proportion value than water?

Error.
It's actually the other way around.
It should read:
Milk to water:227:133
Both are over 120 (not relevant)

Karmanaut:

Error.
It's actually the other way around.
It should read:
Milk to water:227:133
Both are over 120 (not relevant)

Thanks so much!

Pls, I need help on where I can download Advanced Engineering Mathematics by H.K Dass


Thank You

AnnieOakley:
In a primary school, 70% of the boys and 55% of the girls can ride a bicycle. If a boy and a girl are chosen at random, what is the probability that:


(a) both of them can ride a bicycle [Is it 70/100 * 55/100 or 70/200 * 55/199 or neither?]

(b) neither of them can ride a bicycle [Is it 30/100 * 45/100 or 1 - p(both can ride a bicycle)?]

(c) if 150 boys are selected randomly in the school, what is the probability that they can ride a bicycle? [I did 70/100 * 150 = 105. p(both can ride = 105/150 = 7/10). Is it correct?].

Please help @jackpot, laplacian, antoinne, seamione

I don't totally agree with your answer on your previous question.
Maybe you should have considered the 28% of the acid mixture attached to the 300ml
something like this
Ratio of both acid mixture: 2 : 4
Hence you get
2/6(300*28/100)
And
4/6(300*28/100)

NB:The total shouldn't up to 300ml since the acid of the total mixture is just 28%
(from a chem engr approach of mat bal tho)


but for this;
(a)Total % of students=100%

Total% of boys who can ride=70%

Total % of girls who can ride=55%
Thus
Applying the relation
P(AUB)=p(A) + p(B) - P(AnB)

p(AnB)=0.25 or 25%

(b)for neither;
Use 1-(probability of both)
hope it helps

Greetings.

A man spent 1/4 of his monthly salary on rent, 2/5 on food and 1/6 on his children's education. What fraction of his salary is left.?



Show all your works.

Misanthrope:
A man spent 1/4 of his monthly salary on rent, 2/5 on food and 1/6 on his children's education. What fraction of his salary is left.?



Show all your works.

Total fraction =1.
Add all the fractions and subtract from 1 to get the answer.

2nioshine:

I don't totally agree with your answer on your previous question.
Maybe you should have considered the 28% of the acid mixture attached to the 300ml
something like this
Ratio of both acid mixture: 2 : 4
Hence you get
2/6(300*28/100)
And
4/6(300*28/100)

NB:The total shouldn't up to 300ml since the acid of the total mixture is just 28%
(from a chem engr approach of mat bal tho)


but for this;
(a)Total % of students=100%

Total% of boys who can ride=70%

Total % of girls who can ride=55%
Thus
Applying the relation
P(AUB)=p(A) + p(B) - P(AnB)

p(AnB)=0.25 or 25%

(b)for neither;
Use 1-(probability of both)
hope it helps

Why add when the question says 'both'?

AnnieOakley:

Why add when the question says 'both'?

where exactly?.....specify pls

Great tread!!!
















Thumb up guys

2nioshine:



where exactly?.....specify pls

I mean I was expecting something like p(A) * p(B) in the working.

any whatsapp number/ group for this indispensable thread need helps, taking Utme 2016, (complex probability& integration & permutation and combination) Any number please? Mine is 0708-507-8800

hmmmm greetings

#justing.passing.by

agentofchange1:
hmmmm greetings
#justing.passing.by

Lol. Ur signature is funny.
o posted a question in that maths thread. you guys didn't solve it.
diarisgodo

In his pocket a boy has 3 red marbles, 4 blue marbles and 4 green marbles. How many will he have to take out to ensure that he has taken out at least one of each colour?

I got 9, i.e. 4 blue + 4 green + 1 red.

Correct?



@2nioshine, karmanaut, laplacian, agentofchange1

AYOBAMEE911:
any whatsapp number/ group for this indispensable thread need helps, taking Utme 2016, (complex probability& integration & permutation and combination) Any number please? Mine is 0708-507-8800

Mine's +2348101213004

Olaxtra:








Mine's +2348101213004

is there a group?

AnnieOakley:
In his pocket a boy has 3 red marbles, 4 blue marbles and 4 green marbles. How many will he have to take out to ensure that he has taken out at least one of each colour?

I got 9, i.e. 4 blue + 4 green + 1 red.

Correct?

I neither understand the question nor your answer.
Is it a random sample? Does the boy know the color he's going to pick?
Or is it a question under probability?

Karmanaut:

I neither understand the question nor your answer.
Is it a random sample? Does the boy know the color he's going to pick?
Or is it a question under probability?

The marbles are picked at random.

AnnieOakley:

The marbles are picked at random.

The answer is 9. You're correct.

I have a weird question
If you choose an answer at random, what is the chance that you will be correct?
A) 25%
B) 50%
C) 60%
D 25%

Karmanaut:


The answer is 9.
You're correct.

Thanks. It looked so simple so I began to wonder if something grandiose was expected from me.

AnnieOakley:


Thanks. It looked so simple so I began to wonder if something grandiose was expected from me.

At first I was confused: the questioning didn't say it was a random sampling so I thought, why doesn't the guy just take one of each? That would make the answer 3.

Karmanaut:
I have a weird question
If you choose an answer at random, what is the chance that you will be correct?
A) 25%
B) 50%
C) 60%
D 25%

If the correct answer is 25%, then that gives a 50% chance of correctness. Each of the others have a 25% chance of being correct.

AnnieOakley:


If the correct answer is 25%, then that gives a 50% chance of correctness. Each of the others have a 25% chance of being correct.

It is like a feedback loop with no stopping citerion: 'Inputting' an answer of 50% on the first iteration makes the next iteration return an answer of 25%, which in turn puts the NEXT iteration at 50% etc. etc.