Probability Help!!!!!! (1 Viewer)

[lookoutastroboy]

Hey guys (long time no see and merry xmas),

I've just started yr 10 probability and am finding questions related to the product rule hard to understand (P(AB) = P(A) X P(B)) .

Here are some questions that I need help with:

1. There is a spinner with 12 slots numbered from 1-12. It is spun twice.

What is the probability of scoring:

v. two numbers only one of which is prime and
vi. a total greater than 3

and

2. Jabad and Chris each have a regular pack of cards. Each of them selects a card at random from their pack. Find the probability that:
i. their cards are of different colours
ii. Chris has an 8 and Jabad a club
iii. CHRIS HAS AN 8 OR 9 AND jabad has a spade
iv. Jabad doesnt have a club and CHRIS DOESN't have a court card (Jack, QUEEN OR king).
v. Jabad has a 9 of diamonds and Chris has a dimaond or 9

For these questions, could you please show working out in terms of the product rule and use P(x) stuff .



THANKS IN ADVANCE,
LOOKOUTASTROBOY (p.s soz bout use of caps lock in between, my comp is weird).

 

[suzlee]

I hope this is right... It's been too long since I did probability back in year 9. For some reason my school taught us this chapter a year ahead and didn't teach us any of it in year 10.


1. v)
P(A)=6/12 and P(B)=6/12
therefore P(AB)=1/4

vi)
P(A)=10/12 and P(B)=12/12
therefore P(AB)=5/6

2. i)
P(A)=1/2 and P(B)=1/2
therefore P(AB)=1/4

ii)
P(A)=4/52 and P(B)=13/52
therefore P(AB)=1/52

iii)
P(A)=26/52 and P(B)=13/52
therefore P(AB)=1/8

iv)
P(A)=3/4 and P(B)=10/13
therefore P(AB)=15/26

v)
P(A)=1/52 and P(B)=4/13
therefore P(AB)=1/169



Dunno if that's gonna help you at all but there it is

 

[lookoutastroboy]

Thanks for trying suzlee, but most of these answers are wrong unfortunately :S

Any more help would be appreciated guys and gals =)

 

[lyounamu]

Thanks for trying suzlee, but most of these answers are wrong unfortunately :S

Any more help would be appreciated guys and gals =)

can you at least tell us which are right or wrong? so that I can attempt the ones that are wrong?

 

[alcalder]

Let's have a go then. THe trick is thinking of all the possibilities and then using a probability tree. If you haven't learnt about probability trees, yet, look them up because they make these sorts of things SO easy. (Now, I hope I'm right or I'll look like a right nong.)

1. v)
Primes are 1, 2, 3, 5, 7, 11 (unless they are saying 1 is not a prime).
Then
P(A)=5/12 and P(B)=7/12
Let's say 1 is prime, so P(A) = 6/12 and P(B)=6/12
Where A=number prime
B= number not prime.

The spinner is spun twice. So there are 2 possibilities for the first throw:
A or B
so of we have A the first time, then we need B the second time
if we have B the first time we need A the second time.

SO, P(one number prime on 2 throws) = P(AB) + P(BA)
=P(A) x P(B) + P(B) x P(A)
=1/4 + 1/4 = 1/2

EDIT: They are saying 1 isn't prime.

P(1 prime) = 5/12 x 7/12 + 7/12 x 5/12 = 35/144 x 2 = 35/72

To see this properly, you need to draw yourself a probability tree and then use the produce rule horizontally and add vertically.

vi)
So, if the total is to be greater than 3, it cannot equal 3.

EDIT: OK, if the throw total must be greater than 3, then the P(>3)=1-P(<=3)
This means:
P(<=3) = P(sum 2) + P(sum 3)
= P(1) xP(1) + [P(2) x P(1) + P(1) x P(2)]
=1/12 x 1/12 + 1/12 x 1/12 + 1/12 x 1/12
= 1/48

THUS P(total >3) = 1 - 1/48 = 47/48

MUCH SIMPLER!

2. i)
Again you need a tree.
There are three people and two possibilities for each person. And they each have a pack of cards, so we don't need to worry about Jabad taking a card out of the mix before Chris takes his.

Jabad red (1/2) black (1/2)
Chris red (1/2) black (1/2) red (1/2) black (1/2)

red red = 1/4 red black = 1/4 red black = 1/4 black black = 1/4

Thus, P(diff colours) = 1/4 + 1/4 = 1/2

ii)
Chris 8 (1/13) no 8 (12/13)
Jabad club (1/4) no club (3/4) don't really care

P(8 and club) = 1/13 x 1/4 = 1/52

iii)
Chris 8 (1/13) 9 (1/13) neither (11/13)
Jabad spade (1/4) no sp(3/4) spade (1/4) no sp(3/4) don't care

P(8 or 9 & spade) = 1/13 x 1/4 + 1/13 x 1/4 = 1/26

iv)
Jabad club(1/4) no club(3/4)
Chris don't care court card(12/52) no court card(40/52)

P(no club & no court card) = 3/4 x 40/52 = 15/26

v)

EDIT:
OOPS, FORGOT the 9 diamonds had been counted TWICE.

Jabad 9dia (1/52) no 9dia(51/52)
Chris 9dia(1/52) dia other(12/52) 9other(3/52) neither(36/52) don't care

P(9dia & (9 or dia)) = 1/52 x 1/52 + 1/52 x 12/52 + 1/52 x 3/52 = 1/676 + 1/208
= 1/2704 + 12/2704 + 3/2704
= 16/2704 = 1/169

 

[lookoutastroboy]

Here are the solutions:

1.
v. 35/72
vi. 47/48

2.
i. 1/2
ii. 1/52
iii. 1/26
iv. 15/26
v. 1/169



For the questions, could you please keep to the product rule in each as the textbook has specifically (Mathscape yr 10 ext) made each one aligned to it.





Thanks again for the responses, lookoutastroboy

 

[lyounamu]

Here are the solutions:

1.
v. 35/72
vi. 47/48

2.
i. 1/2
ii. 1/52
iii. 1/26
iv. 15/26
v. 1/169



For the questions, could you please keep to the product rule in each as the textbook has specifically (Mathscape yr 10 ext) made each one aligned to it.





Thanks again for the responses, lookoutastroboy

1.

a) 5/12 x 7/12 x 2 = 35/72

b) total number = 144, number of total greater than 3 = 144 - (3) = 141
so p(greater than 3) = 141/144 = 47/48

2a. p(diffrent colour) = 1/2 x 1/2 x 2 = 1/2
b) p(8 and club) = 4/52 x 13/52 = 1/52
c) p(8 or 9 and club) = 8/52 x 13/52 = 1/26
d) p = 3/4 x 40/52 = 15/26
e) p = 1/52 x (12+4-1)/52 = 1/169

 

[lookoutastroboy]

lyounamu, why is his 139/144 question wrong? the working looks alright but could you explain why?














ty,

 

[lyounamu]

lyounamu, why is his 139/144 question wrong? the working looks alright but could you explain why?














ty,

3 possible combi that doesn't equal or come greater than 3, namely (1,1) (1,2) and (2,1) out of 144 combinatinos

so the answer is 141/144

BTW, let me read his working out over again.


he should have written his like this:

1/12 x 10/12 + 1/12 x 11/12 + 10/12 = 47/48

He missed some combinations in his working out.

 

[lookoutastroboy]

last thing if you can,

could you show in detail the working out of the last question 2 e please?
















ty in advance, lookoutastro

 

[lyounamu]

last thing if you can,

could you show in detail the working out of the last question 2 e please?
















ty in advance, lookoutastro

Q. Jabad has a 9 of diamonds and Chris has a dimaond or 9

Chance of a 9 of diamonds = 1/52
Chris has a diamond or 9 = (number of a diamond + chance of a 9 number of 9 and diamond together)/52 = (13+4 -1)/52 = 16/52 = 4/13

if you look at ther card deck, you will see that there is a 9 among diamond cards so we see the overlap and thats why I took 1 away from the total.

so, 4/13 x 1/52 = 1/169

 

[alcalder]

he should have written his like this:

He missed some combinations in his working out.

He is a she!! Please check the gender.

And, yes I made a boo boo, sorry. It would have been easier to do the 1-P(A) anyway. At least I got more right than the last chap

Thanks for the answers. My answers are fixed now.

And, you can't just stay exclusively to the Product Rule. Other rules must come into play - addition, reciprocal etc. And even if a tree is not the way it tells you to solve it, using a tree will not be the end of the world

 

[lyounamu]

He is a she!! Please check the gender.

And, yes I made a boo boo, sorry. It would have been easier to do the 1-P(A) anyway. At least I got more right than the last chap

Thanks for the answers. My answers are fixed now.

And, you can't just stay exclusively to the Product Rule. Other rules must come into play - addition, reciprocal etc. And even if a tree is not the way it tells you to solve it, using a tree will not be the end of the world

Ah sorry! How can I possibly miss that!!!

And agreed. We cannot just stick to product rule. There are so many rules that should be used in conjuction or separately becasue they are so much easier methods...