Probability Questions! (1 Viewer)

[darlking]

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the probability that:
(a) 1 is Faulty
(b) None are Faulty
(c) All 3 cars are faulty

Can you show working out too? Its under tree diagrams but you can just use fractions
Thanksss

 

[lolJK]

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the probability that:
(a) 1 is Faulty
(b) None are Faulty
(c) All 3 cars are faulty

Can you show working out too? Its under tree diagrams but you can just use fractions
Thanksss

EDIT: **incorrect answer

 

[Shadowdude]

So probability of choosing a faulty car is 3/100.

a. 1 is faulty

Ans: 3*(3/100)*(97/100)^2

Or if you do 3u: C(3,1)*(3/100)*(97/100)^2

b. None are faulty

Ans: (97/100)^3

c. All three are faulty

Ans: (3/100)^3


That should be it... I think?

 

[darlking]

P(1 is faulty) = 3/100
P(none are faulty) = 97/100 + 96/99 + 95/98
P(all are faulty) = 3/100 + 2/99 + 1/98

think thats right...

Sorry but its not. Answers are a) 84681/1000000
b) 912673/1000000
c) 27/1000000>>>>>> I figured this one out but the not rest- a and b

 

[Shadowdude]

P(1 is faulty) = 3/100
P(none are faulty) = 97/100 + 96/99 + 95/98
P(all are faulty) = 3/100 + 2/99 + 1/98

think thats right...

Your answer to a can't be right because you have to account for the other two being faulty. You pick three cars, not one.

For b and c, you pick three cars at once, not three cars in succession. And you don't add probabilities because they're not separate cases.

 

[darlking]

So probability of choosing a faulty car is 3/100.

a. 1 is faulty

Ans: 3*(3/100)*(97/100)^2

Or if you do 3u: C(3,1)*(3/100)*(97/100)^2

b. None are faulty

Ans: (97/100)^3

c. All three are faulty

Ans: (3/100)^3


That should be it... I think?

YOUR RIGHT ! How'd you do it, can you explain it to me? first and 2nd and whats the C?

 

[Shadowdude]

YOUR RIGHT ! How'd you do it, can you explain it to me? first and 2nd and whats the C?

Do you do Ext. 1 Maths?


If not, well you make your tree diagram with two choices: faulty or not faulty, with 3/100 or 97/100 probability respectively.

Now if you do that three times, that represents your random selection of three cars.

So if you denote faulty by F and not faulty by N, you'll get things like:

FFF
FFN
FNF
FNN
NFF
NFN
NNF
NNN

or whatever they are and so on...


Count the instances where you select two Not faulty cars and one Faulty car - there are three of them: FNN, NFN, NNF. Add their probabilities (because they are separate cases) - but they're all the same, so it's 3 times the probability of one of them, so 3*(3/100)*(97/100)^2 .

For b, it's just if you pick NNN.


So that's the tree diagram interpretation.

 

[darlking]

Do you do Ext. 1 Maths?


If not, well you make your tree diagram with two choices: faulty or not faulty, with 3/100 or 97/100 probability respectively.

Now if you do that three times, that represents your random selection of three cars.

So if you denote faulty by F and not faulty by N, you'll get things like:

FFF
FFN
FNF
FNN
NFF
NFN
NNF
NNN

or whatever they are and so on...


Count the instances where you select two Not faulty cars and one Faulty car - there are three of them: FNN, NFN, NNF. Add their probabilities (because they are separate cases) - but they're all the same, so it's 3 times the probability of one of them, so 3*(3/100)*(97/100)^2 .

For b, it's just if you pick NNN.


So that's the tree diagram interpretation.

Thank you, your a life saver!

 

[darlking]

In a certain poll, 46% of people surveyed liked the current government, 42% liked the opposition, and 12% had no preference. if two people from the survey are selected at random, find the probability that
A) both will prefer the opposition
B) one will prefer the government and the other will have no preferences
C) both will prefer the government
I need more help, guys

 

[Shadowdude]

In a certain poll, 46% of people surveyed liked the current government, 42% liked the opposition, and 12% had no preference. if two people from the survey are selected at random, find the probability that
A) both will prefer the opposition
B) one will prefer the government and the other will have no preferences
C) both will prefer the government
I need more help, guys

Tree diagram. (with three prongs instead of two)

 

[darlking]

Tree diagram. (with three prongs instead of two)

I did, it looked weird and wrong...

 

[darlking]

I figured out A) Its P(OO)= 0.42^2 = 17.64%

 

[darlking]

I have working out for myself and onlookers I got external aid and I shall post it. This Q. is under tree diagrams but i really fail at them.
For B) P( 1 gov. or none)= P(GN) + P(NG)
= (0.42*0.12) x 2 <--cos two situations and answer is 11%


C) P(GG) = 0.46^2
= 21.2%

Enjoy.