What does it mean first 2 flips are heads

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Homework Help Overview

The discussion revolves around understanding the interpretation of a probability problem involving coin flips. The original poster seeks clarification on what it means for the "first 2 flips to be heads" in the context of tossing a coin six times.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the phrase "first 2 flips are heads" implies specific sequences like {HHTTT} or allows for variations such as {HHTHH}. There are questions about whether the outcomes of the remaining flips affect this interpretation.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the meaning of the problem statement. Some have provided calculations related to the probability of the first two flips being heads, while others are seeking further clarification on the implications of different sequences.

Contextual Notes

There is a mention of a related probability problem involving a die roll, which adds complexity to the discussion. Participants are considering how the outcomes of the coin flips and the die roll interact in determining probabilities.

zeion
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what does it mean "first 2 flips are heads"

Homework Statement



I have a problem that asks me the find the probability that the first 2 flips after tossing 6 coin is heads.

does it mean that it is the result {HHTTTT}?
or does it include results like {HHTHH}, {HHTTH}, etc?


Homework Equations





The Attempt at a Solution

 
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I'm not 100% clear on what your question is. If the question is asking what are the chances that you will have two instances of heads in your first two flips out of 6 flips you simply need to do:

[itex]\frac{1}{2}[/itex] * [itex]\frac{1}{2}[/itex]
(50% chance of heads for the first flip) (multiplied) (50% chance of heads for the second flip)

= [itex]\frac{1}{4}[/itex] = 25%
 


i'm asking, if

{HHTTT} means "first 2 are heads"

then
does

{HHTHH} also mean "first 2 are heads" ?

or does it mean that the rest of the 4 flips must all be tails ?
 


zeion, if we are to interpret what it says from your paraphrasing - and you don't know what it means, how are we to give the correct interpretation?

Write out the question exactly.
 


Suppose that we flip a coin six times and roll a 6-sided die once. Suppose also that all outcomes of this experiment (consisting of an ordered sequence of results for the flips (heads of tails) and the number showing on the die after the roll (an integer from 1 to 6)) are equally likely.

Find the following probabilities:

1) .. i am done
2) The probability that the first R flips are heads, where R is whatever number is showing on the die.
3) ..
4) ..
 


zeion said:
2) The probability that the first R flips are heads, where R is whatever number is showing on
Ok so,
-if the die shows a 2, what is the probability that the first 2 coin flips are heads?
and
-if the die shows a 5, what is the probability that the first 5 coin flips are heads?
and
-etc.
 


so I am asking:

what does it mean first 2 coins are heads?

Is it {HHTTTT} ?
 


zeion said:
so I am asking:

what does it mean first 2 coins are heads?

Is it {HHTTTT} ?
We don't care about the last four flips, so it would be HHXXXX, where X could be H or T.
 


so probability for first 2 are heads is 2^4 / 2^6 ?
 
  • #10


zeion said:
so probability for first 2 are heads is 2^4 / 2^6 ?
Probability that first two flips are heads is .5 x.5 = .25
 
  • #11


oh that is much simpler

so first 2 are heads and roll a 2 is (1/4) * (1/6) = 1/24 ?
 
  • #12


zeion said:
oh that is much simpler

so first 2 are heads and roll a 2 is (1/4) * (1/6) = 1/24 ?

I am having trouble parsing that sentence. Can you please be a little more verbose? Caps and punctuation are required by board rules to aid comprehension.
 
  • #13


Oh, that is much simpler!

Let A be the event that the dice lands on 2.
Let B be the event that the first 2 flips out of 6 are heads.
Let E be the event that the dice lands on 2 and the first 2 flips out of 6 are heads.

Then,
E = A∩B since A and B are independent.

Then,
P(E) = P(A∩B) = P(A)P(B) = (1/6)(1/4) = 1/24

Does that look right?
 

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