ekjchale#1
Active Member
- Joined
- Apr 4, 2022
- Messages
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- HSC
- 2022
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im getting the correct answer with ur method, but can u explain y u did that cuz i assumed if we would do nCr*(p^r)*(q^(n-r)) and as per that i did 20C0 0.8^0 0.2^20. thx in advSo make p = 0.8 (success) and make q = 0.2 (failure)
total sample space is 20,
so for a) i.
i assume you do 20C0 (p)^20 * (q)^0 ?
for ii) 20C1 (p)^19 * (q)^1 ?
and so on.
Please let me know if im correct as its been some time since i've done these questions.
If i'm wrong and you figure out how to do the question pls tell me.
i think cuz part a) is asking for number of FAILURES as it says 'number of boxes that DO NOT contain 47 matches'. So if 0 boxes DO NOT contain 47 matches, that means all 20 boxes contain 47 matches and hence it will be p^20 and q^0.im getting the correct answer with ur method, but can u explain y u did that cuz i assumed if we would do nCr*(p^r)*(q^(n-r)) and as per that i did 20C0 0.8^0 0.2^20. thx in adv
aye that makes sense thx bro, gotta read the questions properly to avoid these silly mistakesi think cuz part a) is asking for number of FAILURES as it says 'number of boxes that DO NOT contain 47 matches'. So if 0 boxes DO NOT contain 47 matches, that means all 20 boxes contain 47 matches and hence it will be p^20 and q^0.
For 1 box not having 47 matches, that means 19 boxes are successes and 1 is a failure. So p^19 and q^1.