What is the probability of obtaining a head and a six when a fair coin and and a die are tossed together?
\(\frac{1}{12}\)
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(\frac{2}{3}\)
Correct answer is D
\(\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}\)
= \(\frac{1}{2} + \frac{1}{6} = \frac{2}{3}\).
Hint: Probability of A and B occurring should be greater than probability A or B happening.