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#1 2024-05-15 10:19:48
- mathxyz
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- From: Brooklyn, NY
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P(at least 1 heads)
Probability of getting at least 1 “heads” on 5 coin flips.
Let me see. I understand it this way.
P(5 coin flips) = (1/2)^5 = 1/32.
P(at least one head) = 1 - P(5 coin flips)
P(at least one head) = 1 - (1/32)
Is this correct?
Last edited by mathxyz (2024-05-15 10:34:04)
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#2 2024-05-15 19:51:24
- Bob
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- Registered: 2010-06-20
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Re: P(at least 1 heads)
Right answer.
P(5 coin flips) = (1/2)^5 = 1/32.
Strictly this ought to be P(no heads) = 1/32
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob 
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#3 2024-05-16 01:42:40
- mathxyz
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- From: Brooklyn, NY
- Registered: 2024-02-24
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Re: P(at least 1 heads)
Right answer.
P(5 coin flips) = (1/2)^5 = 1/32.
Strictly this ought to be P(no heads) = 1/32
Bob
Oh, ok. Thanks for the correction. Is there a formula for AT LEAST probability problems?
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#4 2024-05-16 02:15:17
- Bob
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Re: P(at least 1 heads)
Well sort of. In this example
P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) + P(no heads) = 1
It has to be 1 because one of those events must happen.
If we call P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) the same as P(some heads) then we have
P(some heads) + P(no heads) = 1 and that's what you are using.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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#5 2024-05-16 06:57:45
- mathxyz
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- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053
Re: P(at least 1 heads)
Well sort of. In this example
P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) + P(no heads) = 1
It has to be 1 because one of those events must happen.
If we call P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) the same as P(some heads) then we have
P(some heads) + P(no heads) = 1 and that's what you are using.
Bob
It is more involved that I thought.
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