If Robbie tosses a coin 8 times, what are the odds that at least twice it will be heads??
i know i have to do
(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)
and then multipy but how would i figure out the odds of 2 heads
The probability of at least two heads is 1 - (probability of less than two heads = 1 - (probability of 1 head + probability of no heads) $\displaystyle = 1 - \left[ 8 \cdot \left( \frac{1}{2} \right)^8 + \left(\frac{1}{2}\right)^8\right] = \, .... $.
Now use the definition of odds ... (perhaps see this thread: http://www.mathhelpforum.com/math-he...y-problem.html)