Let
B denote a van with brake problem.
Let abcdef be the event that van with characteristic a, b, c, d, e and f are chosen in this order.
Note that there are two types of vans- one with characteristic B (2 in number) and another with characteristic Bc. The total number of events is the number of ways one choose 2 positions when 6 positions are available, i.e. C26=15
b. Let M be the event that no more than four vans need to be tested before both brake problems are detected.
The number of favourable cases would be equal to the number of ways one can 2 positions when 4 positions are available, i.e. C24=6.
\begin{align*} P(M)&=\dfrac{\textrm{No. of favourable events}}{\textrm{Total number of events}} \\ &=\dfrac{6}{15}\\ &=0.4 \end{align*}P(M)=Total number of eventsNo. of favourable events=156=0.4
Answer
a.
0.2 b.
0.4 c.
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