TYS 'O' LVL maths question

tinuviel07 SG

it's like

1/T^2 =5

you bring T^2, does it become T^1/2?

it becomes

1 = 5 x T^2,

the divide becomes multiply. doesn't affect the root

gunner77 SG

ohhh i get it

thanks tinuviel07!

maths pro

thanks for the help

tinuviel07 SG

sure no prob.. see ya

MrSean SG

sq root bring over is not sq meh?

MUST... WIN....

Well gunner you could square it if you want.

(T^4/T^0.5)^2 = (T^n)^2

T^8/T^1 = T^2n

T^7 = T^2n

7=2n

n = 3.5

Crap lost again.

gunner77 SG

see ya

tmr i post more

tinuviel07 SG

MUST... WIN....

Well gunner you could square it if you want.

(T^4/T^0.5)^2 = (T^n)^2

T^8/T^1 = T^2n

T^7 = T^2n

7=2n

n = 3.5

Crap lost again.

haha pat pat

gunner77 SG

MUST... WIN....

Well gunner you could square it if you want.

(T^4/T^0.5)^2 = (T^n)^2

T^8/T^1 = T^2n

T^7 = T^2n

7=2n

n = 3.5

Crap lost again.

wah bro

tis one very hard to understand

gunner77 SG

can i ask one last question?

MrSean SG

Please ask so I have a chance of winning.... Perhaps he has gone to bed and I will have an open field...

tinuviel07 SG

i'll leave the field to you. bye..

gunner77 SG

(A) two unbiased dice are thrown.

find the probability that they

(i) show the same number

(ii) show different number

(B) three unbiased dice are thrown.

find the probability that

(i) they all show different numbers

(ii) at least two show the same number

please kindly explain how u guys get the ans.

MrSean SG

Ah well.

tinuviel07 SG

(A) two unbiased dice are thrown.

find the probability that they

(i) show the same number

(ii) show different number

(B) three unbiased dice are thrown.

find the probability that

(i) they all show different numbers

(ii) at least two show the same number

please kindly explain how u guys get the ans.

i suck at probability

(ai) 1/6,

do tree diagrams or whatever you want to do to find the scenarios like

(1,1) , (2, 2), (3, 3) and so on.. each scenario probability = 1/36,

there's 6 of these scenarios so 6 x 1/36 = 1/6

(btw it's 1/36 + 1/36 + 1/36 ...)

(aii) ans = 1 - ans in (ai) = 1 - 1/6 = 5/6

gunner77 SG

actually part A i get correct

part B is the problem

but i wan to clarify something in A(i)

2 dice thrown rite, so we have to find the probability both dice show the same number.

i thought its 1/6 for dice 1 then 1/6 for dice 2

add up

1/6 + 1/6 = 2/6

= 1/3 ??

why is tis wrong?

gunner77 SG

btw the answer is

a(i) 1/6

a(ii) 5/6

b(i)5/9

b(ii)4/9

tinuviel07 SG

u know how to draw tree diagram? easier for you to see

1/6 is probability of throwing any no. of that dice.. doesn't describe the scenario accurately..

gunner77 SG

ohh ok, i get it why im wrong now

its actually 1/6 X 1/6

how to do B ah?

tinuviel07 SG

yup

eagle SG

(A) two unbiased dice are thrown.

find the probability that they

(i) show the same number

(ii) show different number

(B) three unbiased dice are thrown.

find the probability that

(i) they all show different numbers

(ii) at least two show the same number

please kindly explain how u guys get the ans.

(A) (i) P(both shows same number) = 1/6 * 1/6 * 6 (6 different numbers) = 1/6

(ii) P(both shows different number) = 1 - both show same number = 5/6

(B) (i) P(all different numbers) = P(first 2 dice are different and 3rd dice is also different number)

= 5/6 * 4/6 (4 numbers out of 6 left)
= 5/9

(ii) P (at least two show same number) = 1 - P(all different numbers) = 4/9

Question will be placed in ExamWorld soon :D

gunner77 SG

(A) (i) P(both shows same number) = 1/6 * 1/6 * 6 (6 different numbers) = 1/6

(ii) P(both shows different number) = 1 - both show same number = 5/6

(B) (i) P(all different numbers) = P(first 2 dice are different and 3rd dice is also different number)

= 5/6 * 4/6 (4 numbers out of 6 left)
= 5/9

(ii) P (at least two show same number) = 1 - P(all different numbers) = 4/9

Question will be placed in ExamWorld soon :D

bro u get correct

but can u explain how u do part B?

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