A coin is tossed and a die is thrown. The probability that the outcome will be head or a number greater than…
A coin is tossed and a die is thrown. The probability that the outcome will be head or a number greater than 4 or both, is
$\frac{2}{3}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{3}$
Solution
The probability of getting head when a coin is thrown $=\frac{1}{2}$ The probability of getting a number greater than 4 when a die is thrown
$=\frac{2}{6}=\frac{1}{3}$
Hence required probability $=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{4}{6}=\frac{2}{3}$