Question on: WAEC Mathematics - 2005
A fair die is tossed once, what is the probability of obtaining neither 5 or 2
A
\(\frac{5}{6}\)
B
\(\frac{2}{3}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{6}\)
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Correct Option: B
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 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)
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