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04 Aug 2014, 08:13
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Hi guys, by this time all I've done was read read read... but now I have to write, cause I can't find logical solution to this problem.
Have been reading "Probability tutorial" by Walker here on forum, and came across this:
Mutually exclusive events
Q: If Jessica rolls a die, what is the probability of getting at least a "3"?
Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is:
P = 1/6+ 1/6 + 1/6 + 1/6 = 2/3
Meanwhile
In "Kaplan premier 2013" chapter16, p814
Same principal is being tested but the way of getting there is cosmically different:
If a fair coin is flipped three times, what is the probability of getting at least one tail ?
Suggested way of tackling this one is:
Total - Undesired
1 - HHH
1- 1/2*1/2*1/2
1-1/8
= 7/8
Please advise
edit: my question is, is there a way of doing both ways in dice case as well as in coin case ?
Have been reading "Probability tutorial" by Walker here on forum, and came across this:
Mutually exclusive events
Q: If Jessica rolls a die, what is the probability of getting at least a "3"?
Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is:
P = 1/6+ 1/6 + 1/6 + 1/6 = 2/3
Meanwhile
In "Kaplan premier 2013" chapter16, p814
Same principal is being tested but the way of getting there is cosmically different:
If a fair coin is flipped three times, what is the probability of getting at least one tail ?
Suggested way of tackling this one is:
Total - Undesired
1 - HHH
1- 1/2*1/2*1/2
1-1/8
= 7/8
Please advise
edit: my question is, is there a way of doing both ways in dice case as well as in coin case ?
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04 Aug 2014, 13:52
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realist
I will show you how to solve in both direction. Just remember that in probability you can solve in both directions, but one of them always faster. So, just choose the correct one!
Mutually exclusive events
Total - Undesired
There are 2 outcomes that don't satisfy our condition (at least 3): {1, 2}. The probability of each outcome is 1/6. The probability of getting at least a "3" is:
P = 1-(1/6+ 1/6) = 2/3
Meanwhile
There are 7 outcomes out of 8 that satisfy our condition (getting at least 1 tail): THH, HTH, HHT, TTH, THT, HTT, TTT (actually you can say here that probability is 7/8)
By Kaplan the corresponding probability is
1/2*1/2*1/2+1/2*1/2*1/2+1/2*1/2*1/2+1/2*1/2*1/2+1/2*1/2*1/2+1/2*1/2*1/2+1/2*1/2*1/2=7/8
Now we see that in both problem it is better go through "undesired" outcomes.
04 Aug 2014, 20:56
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Thanks for this answer, but why are we multiplying in coin case and adding in dice case?
Is there a reason for that? Because as I see it, it is the same exclusive/independent action... If you cast a dice once, its the same as pulling a card once.
Posted from my mobile device
Is there a reason for that? Because as I see it, it is the same exclusive/independent action... If you cast a dice once, its the same as pulling a card once.
Posted from my mobile device
