Answer:
\(\frac{1}{3}\)
Step-by-step explanation:
Total number of events on dice = {1,2,3,4,5,6}=6
Favorable event = {3}=1
So, probability of getting 3 on dice = \(\frac{\text{Favorable events }}{\text{Total events}}\)
= \(\frac{1}{6}\)
Favorable event = {4}=1
So, probability of getting 4 on dice = \(\frac{\text{Favorable events }}{\text{Total events}}\)
= \(\frac{1}{6}\)
So, probability of getting 3 or 4 on a dice = \(\frac{1}{6}+\frac{1}{6}\)
= \(\frac{2}{6}\)
= \(\frac{1}{3}\)
Hence P(3 or 4) is \(\frac{1}{3}\)
