TYS 'O' LVL maths question
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Originally posted by MrSean:
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
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Originally posted by MrSean:
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
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can i ask one last question?
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Please ask so I have a chance of winning.... Perhaps he has gone to bed and I will have an open field...
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i'll leave the field to you. bye..
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(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.
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Ah well.
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Originally posted by gunner77:
(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
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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?
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btw the answer is
a(i) 1/6
a(ii) 5/6
b(i)5/9
b(ii)4/9
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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..
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ohh ok, i get it why im wrong now
its actually 1/6 X 1/6
how to do B ah?
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yup
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Originally posted by gunner77:
(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
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Originally posted by eagle:
(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 correctbut can u explain how u do part B?
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draw tree diagram ba.. i got to slp liao.. wait for mr sean to rescue you
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Originally posted by gunner77:
bro u get correctbut can u explain how u do part B?
part B (i) get from part A (ii)
if the first two dice are already different (5/6 probability), the probability of the 3rd dice having a different number from the first two dice is 4/6
Hence 5/6 * 4/6
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Originally posted by tinuviel07:
draw tree diagram ba.. i got to slp liao.. wait for mr sean to rescue you
good nite
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zzzz this kind of question really no cake liao lor for me.. gun gun still got long way
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Originally posted by eagle:
part B (i) get from part A (ii)
if the first two dice are already different (5/6 probability), the probability of the 3rd dice having a different number from the first two dice is 4/6
Hence 5/6 * 4/6
i tink i roughly get it lathanks!
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Originally posted by youyayu:
zzzz this kind of question really no cake liao lor for me.. gun gun still got long way
no cake means wat?too easy for u ah?
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hey TIN TIN
i got a question.
factorise fully
4ax^2 - 3b + 12bx^2 - a
3 marks question.
how to do sia?
ans is (a+3b)(2x+1)(2x-1)
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how you know i online..
= (12bx^2-3b) + (4ax^2-a)
= 3b(4x^2-1)+a(4x^2-1) take out common factor
= (3b+a)(2x+1)(2x-1)
your answer correct ah
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Originally posted by tinuviel07:
how you know i online..
= 3b(4x^2-1)+a(4x^2-1)
= (3b+a)(2x+1)(2x-1)
= (3b+a) (4x2 - 1)
your answer correct ah
the answer is behind the TYS.at the bottom of the screen got write 50 user online wat.
can see ur username
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eh sorry sorry
ur ans correct!
typo