Compute the odds in favor of obtaining?
At least 1 head when a single coin is tossed 3 times?
help me please i know the answer is 7:1 but how and what did they do to get that answer?
It is certain that for three tosses that there is either no heads (event X) or at least 1 head (event Y). Thus:
$\displaystyle P(X)+P(Y)=1$
$\displaystyle P(Y)=1-P(X)$
Now, what it the probability of no heads (3 tails)?
What I mean is, I am asking you to compute the probability that all three tosses are tails. Once you do this, then you can complete the question.
What is the probability that for 1 toss, the result is tails?
No, the probability of getting the first tail AND getting the second tail AND getting the third tails is:
(1/2)(1/2)(1/2) = 1/8
You see, there are 8 outcomes, only one of which is 3 tails:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Yes, you can see now, even without the formula that there are 7 outcomes with at least 1 heads and only 1 without. The formula is useful though...for example if instead of 3 tosses there are 100 and listing the 2^100 outcomes is just not practical.