Section II (2 Viewers)

[SMS.EDWIN]

27(c) i got even chance aswell. Got to last page knowing i didnt hav enough time. So spent it working out 28(c) then after the exam realised i got the fraction the other way around and got like two thousand metres or something. FML.

 

[susie22]

there cjosie16

But what about her chance of winning BOTH prizes?

 

[2pmhottest]

Guys for Question 27 C GUYSMary has the better chance of winning let me explain this to youMary has 2/100 tickets which is 1/50 which is 2 percent. Jane has 1/100 + 1/100 so she also has 2 percent chanceHOWEVER the question states determine the better chacne of winning AT LEAST one PRIZEso there for mary still has a 2/100 chance of winning at least one prize and also 98/100 to LOSEJANE HOWEVER has 1/100 to win 1st raffle and 99/100 to lose the 2nd raffle is the same 1/100 to win and 99/100 to loseBUT JANE U MUST CALCULATE HER CHANCE OF WINNING AT LEAST 1 prize so it is1/100 x 99/100 + 1/100 x 99/100 WHICH EQUAL to 1.98% of her winningso therefore MARY HAS a 2 % chance of winning while jane has a 1.98 chance of WINNING AT LEAST ONE PRIZEand i am 99% sure this is the correct answer let me know what u guys think</SPAN> .............OK THERE IT IS !!!</SPAN>

 

[hazza9]

Guys for Question 27 C GUYSMary has the better chance of winning let me explain this to youMary has 2/100 tickets which is 1/50 which is 2 percent. Jane has 1/100 + 1/100 so she also has 2 percent chanceHOWEVER the question states determine the better chacne of winning AT LEAST one PRIZEso there for mary still has a 2/100 chance of winning at least one prize and also 98/100 to LOSEJANE HOWEVER has 1/100 to win 1st raffle and 99/100 to lose the 2nd raffle is the same 1/100 to win and 99/100 to loseBUT JANE U MUST CALCULATE HER CHANCE OF WINNING AT LEAST 1 prize so it is1/100 x 99/100 + 1/100 x 99/100 WHICH EQUAL to 1.98% of her winningso therefore MARY HAS a 2 % chance of winning while jane has a 1.98 chance of WINNING AT LEAST ONE PRIZEand i am 99% sure this is the correct answer let me know what u guys think</SPAN> .............OK THERE IT IS !!!</SPAN>

I did it similar, pretty much got the same answer

MARY is easy 2/100 = 2%

For JANE I found the probability she would NOT win either – hence leaving me with the probability she would win AT LEAST ONE

I did P(win neither) = 99/100 x 99/100
.............................= 98.01%


Thus, probability she will WIN AT LEAST ONE is 1.99%

MARY = 2%
JANE = 1.99%

 

[mitch47]

wow i did that probability question heaps differently im starting to get worried now lol i drew a tree diagram one each for Mary and Jane and to find the probability of winning AT LEAST ONE i went through the three prize draws and found the combinations of win,lose,win etc.

because with mary my first branch of the first draw was win (2/100) and lose (98/100) so anything on the bottom half had no chance of winning at all whereas with Jane there were more possibilities (eg, WWL, WLW, LWW, LWL, LLW, WLW) there were just more combinations so I put jane

geez I dunno anxious now

 

[sadcase]

I did the same... i drew a tree diagram and found that Mary had a higher chance. For the other girl i got 0.001 something like that. hopefully i get 3 out of 4
lol

 

[Enchantress91]

I don't get it peeps, how can it be 1/100 x 1/100? How can Jane have 0.0001 (or something like that) chance??? When she bought 2 tickets out of 200!

Come on guys, think about it this way, she bought 2 tickets out of 200 with 2 prizes (one in each raffle). What you guys did was youse calculated the probability of Jane winning BOTH prizes, which was not what the Question asked us to find.

Hence Jane has 1/100 chance of winning. Not 0.0001% (1/10000)

I'm 99.9999% sure about this. Ask your math teachers if you're still having trouble... though I know the exam's over...