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secondary 4 | E Maths
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Hi,
could you kindly help to solve part (b), why isn’t the answer 0.0023, but 0.002297?
And part (e) where information will be needed from parts (a) and (b) of the question (how 4D betting system works and the prize structure).
Appreciate your advice, many thanks!
Date Posted: 3 years ago
Views: 654
b)
There are 10000 possible numbers , from 0000 to 9999.
For each prize, there is a 1/10000 (or p = 0.0001) chance that a particular number will be drawn.
But you can't just do 0.0001 × 23 = 0.0023 since it's not the probability.
0.0023 is the probability when you bought 23 different numbers and hoping for them to appear for ONE prize.
Eg. If you bought 1011 to 1023, what is the probability that any of these numbers appear for the first prize?
There are 10000 possible numbers , from 0000 to 9999.
For each prize, there is a 1/10000 (or p = 0.0001) chance that a particular number will be drawn.
But you can't just do 0.0001 × 23 = 0.0023 since it's not the probability.
0.0023 is the probability when you bought 23 different numbers and hoping for them to appear for ONE prize.
Eg. If you bought 1011 to 1023, what is the probability that any of these numbers appear for the first prize?
Here, you have 23 prizes so it's having 23 numbers drawn, one at a time.
Think of it as drawing 1 ball out of a bag containing 10000 uniquely numbered balls, 23 times, with replacement.
Now to calculate the required probability in the forward way, it will be too cumbersome as you'll need to calculate the following :
P(the number wins only 1 prize)
+P(the number wins 2 prizes)
+P(the number wins 3 prizes)
+...
...
+P(the number wins all 23 prizes)
"Any prize" means it could be 1 or more prizes
So, we use the complement method.
P(a particular number wins any prize)
= 1 - P(a particular number does not win any prize at all)
= 1 - (0.9999)²³
[Because probability of a particular number not winning one prize = 9999/10000 i.e it is won by the other 9999 numbers.
Power 23 because there are 23 draws, so it is just 0.9999 × 0.9999 × ... 0.9999 (23 '0.9999's).
the analogy is the ball example mentioned above)
≈ 0.00229747177
= 0.002297 (4s.f)
Think of it as drawing 1 ball out of a bag containing 10000 uniquely numbered balls, 23 times, with replacement.
Now to calculate the required probability in the forward way, it will be too cumbersome as you'll need to calculate the following :
P(the number wins only 1 prize)
+P(the number wins 2 prizes)
+P(the number wins 3 prizes)
+...
...
+P(the number wins all 23 prizes)
"Any prize" means it could be 1 or more prizes
So, we use the complement method.
P(a particular number wins any prize)
= 1 - P(a particular number does not win any prize at all)
= 1 - (0.9999)²³
[Because probability of a particular number not winning one prize = 9999/10000 i.e it is won by the other 9999 numbers.
Power 23 because there are 23 draws, so it is just 0.9999 × 0.9999 × ... 0.9999 (23 '0.9999's).
the analogy is the ball example mentioned above)
≈ 0.00229747177
= 0.002297 (4s.f)
For d) I will interpret it as the number winning both first and second prize only, and that number does not win any other prize.
For this the probability will be :
0.0001 × 0.0001 × 0.9999²¹
[0.0001 are for each of the first and second prizes and 0.9999²¹ is for the other 21 prizes that the number doesn't win]
≈ 9.97902099 x 10⁻⁹
= 9.98 × 10⁻⁹ (3s.f)
However, if we are only concerned about the first 2 prizes and results of the other prizes don't matter (i.e this number could either win or not win them), then it will simply be:
0.0001²
= 0.00000001 (or 1 × 10⁻⁸) (or 1/100 000 000, which is one out of one hundred million)
For this the probability will be :
0.0001 × 0.0001 × 0.9999²¹
[0.0001 are for each of the first and second prizes and 0.9999²¹ is for the other 21 prizes that the number doesn't win]
≈ 9.97902099 x 10⁻⁹
= 9.98 × 10⁻⁹ (3s.f)
However, if we are only concerned about the first 2 prizes and results of the other prizes don't matter (i.e this number could either win or not win them), then it will simply be:
0.0001²
= 0.00000001 (or 1 × 10⁻⁸) (or 1/100 000 000, which is one out of one hundred million)
For e), think of it like this :
There are 10000 choices for the first prize.
There are 10000 choices for the second prize.
Some possibilities :
First prize 1234 , second prize 9999
First prize 9999, second prize 1234
First prize 0000, second prize 0000
First prize 8888, second prize 8888
Total number of permutations
= 10000 × 10000
= 100 000 000
(100 million)
Out of these, you only need the following
First prize 0000, second prize 0000
First prize 0001, second prize 0001
First prize 0002, second prize 0002
.
.
.
First prize 9999, second prize 9999
That's a total of 10000 permutations.
Required probability
= 10000 / 100 000 000
= 1/10000
Or
Simply take the answer in d) and multiply by 10000 since that one is for 1 number, and here in e) you'll need to find for 100000 numbers
1/100 000 000 × 10000
= 1/10000
There are 10000 choices for the first prize.
There are 10000 choices for the second prize.
Some possibilities :
First prize 1234 , second prize 9999
First prize 9999, second prize 1234
First prize 0000, second prize 0000
First prize 8888, second prize 8888
Total number of permutations
= 10000 × 10000
= 100 000 000
(100 million)
Out of these, you only need the following
First prize 0000, second prize 0000
First prize 0001, second prize 0001
First prize 0002, second prize 0002
.
.
.
First prize 9999, second prize 9999
That's a total of 10000 permutations.
Required probability
= 10000 / 100 000 000
= 1/10000
Or
Simply take the answer in d) and multiply by 10000 since that one is for 1 number, and here in e) you'll need to find for 100000 numbers
1/100 000 000 × 10000
= 1/10000
See 1 Answer
See question's comments section for explanation.
Thank you so much for your advice, Mr J.
For part (b), does the cumbersome method of calculation mean 23C1 (0.0001) (0.9999)^22 + 23C2 (0.0001)^2 (0.9999)^21 + …. + 23C22 (0.0001)^22 (0.9999)?
For part (b), does the cumbersome method of calculation mean 23C1 (0.0001) (0.9999)^22 + 23C2 (0.0001)^2 (0.9999)^21 + …. + 23C22 (0.0001)^22 (0.9999)?
Yes. If you try it you'll get the same answer as 1 - 0.9999²³
Note that you can choose to put a 23 C 23 in front of 0.9999²³ if you want to, but it's not really needed since 23 C 23 just equals 1.
i.e there is only one way to not win any prizes.
Note that you can choose to put a 23 C 23 in front of 0.9999²³ if you want to, but it's not really needed since 23 C 23 just equals 1.
i.e there is only one way to not win any prizes.
Thanks Mr J. I got it now :)
The analogy really helped!
I hope you have a good day ahead. Stay safe and take care :D
The analogy really helped!
I hope you have a good day ahead. Stay safe and take care :D
Welcome.
The analogy however, would be too cumbersome for real-life application (10000 differently numbered balls is a lot to manufacture and handle!) so how the numbers are picked for 4D (or equivalent games in the world) is something like this :
There are 4 slots, one for each digit.
The digits are picked one at a time.
For each slot, there are 10 balls numbered from 0 to 9. A machine will spin these 10 balls until one comes out of its opening and falls into the slot.
Similarly for Toto, 6 numbers and 1 additional number are picked out of 49, but for this there is no replacement so the machine only releases 7 numbers, 1 at a time.
The analogy however, would be too cumbersome for real-life application (10000 differently numbered balls is a lot to manufacture and handle!) so how the numbers are picked for 4D (or equivalent games in the world) is something like this :
There are 4 slots, one for each digit.
The digits are picked one at a time.
For each slot, there are 10 balls numbered from 0 to 9. A machine will spin these 10 balls until one comes out of its opening and falls into the slot.
Similarly for Toto, 6 numbers and 1 additional number are picked out of 49, but for this there is no replacement so the machine only releases 7 numbers, 1 at a time.
Thank you so much for your great advice, Mr J.
Really appreciate :)
Really appreciate :)
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