I flip 2 (fair) coins and tell you that at least one of them is heads. What are the odds that both are heads?

Answer 1

I beg to differ. The probabilty is 1/3. Consider the sample space...
You flip 2 coins so the sample space is
first coin, second coin
H,H
H,T
T,H
T,T
Your question, translated into the language of conditional probability, is,
given that at least one is heads, what is the probability that both are heads?
Conditional probability replaces the "given" subset of the sampel space for the entire subset. So the probability breaks down as
number of elements in {H H} / number of elements in {HT TH HH}
That probability is 1 in 3
Odds are a little different from probability, odds can be thought of as the ratio of "won't" to "will"
http://mathforum.org/library/drmath/view/65403.html
So if you know that at least one is heads, then the odds that both are heads is the ratio of
one isn't heads ... to ... both are heads
The number of ways that one isn't heads is 2
The number of ways both are heads is 1
So the odds are 2:1 that both are heads, and the probabliity that both are heads, given that one is a head, is 1/3

Answer 2

This is the same idea as your question about the mother who has at least one son. The simplest interpretation of that statement has a correct answer of 1/3. A different interpretation gives an answer of 1/2.

The simple interpretation is as follows:
The person making the statement flipped 2 coins repeatedly until at least one head came up. He/she then said, "At least one is a head." In that case the answer is 1/3.

A more complex interpretation is:
The person flipped two coins, looked at the first coin, and whatever it was, he/she announced, "At least one is a (whatever that first coin was)." In that case the answer is 1/2.

Answer 3

1/2

Answer 4

To get >= 2 heads, you like: a million) 2 heads (& a million tail), plus 2) 3 heads. risk of two heads & a million tail = a million/2 x a million/2 x a million/2 x 3 (can are available in distinctive mixtures of HHT, HTH, or THH) = 3/8 plus risk of three heads = a million/2 x a million/2 x a million/2 = a million/8 as a result risk of a minimum of two heads = 3/8 + a million/8 = a million/2.

Answer 5

She-nerd is right on this one.

consider the conditional probabillity:

P( both heads | at least one head) =
P( both heads and at least one head)/P( at least one head)=
P( both heads )/P( at least one)=
(1/4)/(3/4) = 1/3

Note that the intersection of events
'both heads'
and
'at least one'
is 'both heads'

Answer 6

These are the only possibilities: hh, ht, th. Therefore, the answer is 1/3.

Answer 7

it is a 50 50 shot

Answer 8

HH
HT

50%