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Mathematics: Post your doubts here!

salvatore

salvatore

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syed1995 said:
Q7 i) Atleast one 3 .. means that it is saying P(X≥1) or P(X>0) n = 9

Where X is the number of 3s.

So for it to be zero 3s the probability will be (5/6) and 0 3s at all 9 throws would be P(X=0)=(5/6)^9

Now it asked at least 1 3 .. meaning there are no zeros .. so 1 - P(X=0) = 1- (5/6)^9 = 0.806 Answer

ii) When n fair dice are thrown, the probability of getting at least one 3 is greater than 0.9.
Find the smallest possible value of n.

This is a real tricky question .. we know that for one dice which is thrown 9 times the probability of getting atleast one 3 is 1-(5/6)^9

so for n number of throws the probability will be 1- (5/6)^n

Now the question says .. the probability of n number of throws is 0.9 so

0.9 = 1-(5/6)^n
-0.1=-(5/6)^n
(5/6)^n=0.1

Now To get the value of n we need to use ln on both sides (a little something from my add maths).. using ln would get the power to the bottom

n* ln (5/6)= ln(0.1)
n = ln (0.1) / ln(5/6)
n = 12.62

so atleast 13 throws need to be made for the probability to be 0.9.

b)

This was a real tricky question .. I remember my friend asked me solve it..

5 Green Balls 3 Yellow Balls .. Player wins when a yellow ball is drawn.. Julie draws the first ball .. Probability of Ronnie Winning..

The ways in which ronnie can win are as follows

J'R , J'R'J'R , J'R'J'R'J'R (then the green balls end so yellow ball would come eventually..) To summarize these details in terms of balls
GY , GGGY, GGGGGY

Total = 9, 5 green and 3 yellow.. so probabilities become

P(GY) = (5/8 * 3/7) = 15/56

P(GGGY) = (5/8 * 4/7 * 3/6 * 3/5) = 3/28

P(GGGGGY) = (5/8 * 4/7 * 3/6 * 2/5 * 1/4 * 3/3) = 1 / 56

Total P = 15/56 + 3/28 + 1/56 = 11/28 Answer


Hope that helps...
Click to expand...
Thanks a lot for your help bruv.. much appreciated.
I don't understand a few things though:

For the first part, why can't I solve it this way: P(X=1) = (1/6) => (1/6)^9? (Since only one 3 out of 6 => 1/6)

For part (ii), is there any other way apart from log to solve for the value of n? I'm not really familiar with logarithm..

Finally, I totally don't understand how you solved part(b).. how did you get GY , GGGY, GGGGGY? Please help me understand this

I'm sorry if I'm bothering you with all these questions, but I really need your help.. I suck at statistics!
 
syed1995

syed1995

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Rutzaba said:
im in uni bro :p
Click to expand...


Nice! Which University?

salvatore said:
Thanks a lot for your help bruv.. much appreciated.
I don't understand a few things though:

For the first part, why can't I solve it this way: P(X=1) = (1/6) => (1/6)^9? (Since only one 3 out of 6 => 1/6)

For part (ii), is there any other way apart from log to solve for the value of n? I'm not really familiar with logarithm..

Finally, I totally don't understand how you solved part(b).. how did you get GY , GGGY, GGGGGY? Please help me understand this

I'm sorry if I'm bothering you with all these questions, but I really need your help.. I suck at statistics!
Click to expand...

First Part .. Well it is saying atleast one 3 .. not only one three .. so we need the probability of P(X≥1) .. it can be 1 three or 2 three or 3 three or even nine 3 it just can't be zero 3's... You need to solve it for X=1,2,3,4,5,6,7,8,9 and then add them all to get the correct answer .. or do what I did.

Second Part .. no other way that i know off .. I usually use my calculator to do these equations .. like i have a fx 991ES .. just do this

(1-(5/6)^x) then [ALPHA] the buttom below shift (It's SOLVE/CALC for me) then type the 0.9 and hit SOLVE [Shift and then press CALC]

so it becomes (1-(5/6)^x) = 0.9 then SOLVE .. It takes a little while .. but it gets the correct value of X.

Honestly there is no other way than log to resolve powers .. if there is then I don't know of the method.

Part b ..

Well it said that Julie was the first to play .. Yellow meant that the person won and if green was drawn another ball was drawn .. and we needed the probability of ronnie winning.

that is only possible is Julie lost every time (Got a green ball) and then Ronnie got a Yellow one (and won and game stops).. so GY .. but since there are 5 green balls .. until those end the game can continue.. so in the next one .. J gets a G and then R gets a G and then again J gets G and then R gets a Y to win and game stops.. and so on until all 5 green balls get used.. This part is tricky to understand .. read the question 2-3 times and then see my working .. you will hopefully understand it.
 
salvatore

salvatore

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syed1995 said:
Nice! Which University?



First Part .. Well it is saying atleast one 3 .. not only one three .. so we need the probability of P(X≥1) .. it can be 1 three or 2 three or 3 three or even nine 3 it just can't be zero 3's... You need to solve it for X=1,2,3,4,5,6,7,8,9 and then add them all to get the correct answer .. or do what I did.

Second Part .. no other way that i know off .. I usually use my calculator to do these equations .. like i have a fx 991ES .. just do this

(1-(5/6)^x) then [ALPHA] the buttom below shift (It's SOLVE/CALC for me) then type the 0.9 and hit SOLVE [Shift and then press CALC]

so it becomes (1-(5/6)^x) = 0.9 then SOLVE .. It takes a little while .. but it gets the correct value of X.

Honestly there is no other way than log to resolve powers .. if there is then I don't know of the method.

Part b ..

Well it said that Julie was the first to play .. Yellow meant that the person won and if green was drawn another ball was drawn .. and we needed the probability of ronnie winning.

that is only possible is Julie lost every time (Got a green ball) and then Ronnie got a Yellow one (and won and game stops).. so GY .. but since there are 5 green balls .. until those end the game can continue.. so in the next one .. J gets a G and then R gets a G and then again J gets G and then R gets a Y to win and game stops.. and so on until all 5 green balls get used.. This part is tricky to understand .. read the question 2-3 times and then see my working .. you will hopefully understand it.
Click to expand...
I'm getting the concept now..

Thanks a lot for giving your time man :)
 
immie.rose

immie.rose

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Rutzaba said:
depends on u... i loved p3 as compared to s1.
ii gave accelerated... and u wont guess.... with s2
got a b :/
and that also after only two months study of s2 :(
Click to expand...
hmm i thought og giving accel and i think s1 is enough of a threat i wouldnt risk with s2 :unsure:. hmm the other grades would have covered up i guess :)
 
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