Jambmaths question:
The probability of a students passing any examination is $\tfrac{2}{3}$. If the student takes three examination. What is the probability that he will not pass any of them
Option A:
$\tfrac{2}{3}$
Option B:
$\tfrac{4}{9}$
Option C:
$\tfrac{8}{27}$
Option D:
$\tfrac{1}{27}$
Jamb Maths Solution:
$\begin{align} & \text{Probability of passing }=\tfrac{2}{3} \\ & \text{Probability of failing }=1-\tfrac{2}{3}=\tfrac{1}{3} \\ & \text{Probability of failing all examination}=P(FFF) \\ & P(FFF)=\tfrac{1}{3}\times \tfrac{1}{3}\times \tfrac{1}{3}=\tfrac{1}{27} \\\end{align}$
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