WAEC - Mathematics (2005 - No. 8)
A fair die is tossed once, what is the probability of obtaining neither 5 or 2
\(\frac{5}{6}\)
\(\frac{2}{3}\)
\(\frac{1}{2}\)
\(\frac{1}{6}\)
ব্যাখ্যা
Probability of obtaining a 5 is P(5)=\(\frac{1}{6}\)
Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\)
Probability of obatining either 2 or 5 = P(2∪5) = \(\frac{2}{6}\)
Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)
Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\)
Probability of obatining either 2 or 5 = P(2∪5) = \(\frac{2}{6}\)
Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)