A manufacturer of air plane parts makes a certain engine that has a probability $p$ of failing on any given flight. There are two planes fitted with this type of engine. One plane has $3$ such engines and other plane has $5$ . A plane crashes if more than half the engines fitted in it fail. If the two planes have the same probability of crashing, then the possible value of $p$ are

\frac{3}{5}

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\frac{1}{2}

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1

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Solution

The correct options are
B $\frac{1}{2}$
C $1$
D $0$
$^{3} C_{2} p^{2} (1 - p) +^{3} C_{3} p^{3} =^{5} C_{3} p^{3} (1 - p)^{2} +^{5} C_{4} p^{4} (1 - p) +^{5} C_{5} p^{5} 3 p^{2} (1 - p) + p^{3} = 10 p^{3} (1 - p)^{2} + 5 p^{4} (1 - p) + p^{5}$
$\Rightarrow p = 0$ and $10 p (1 + p^{2} - 2 p) + 5 p^{2} (1 - p) + p^{3} - 3 (1 - p) - p = 0$
$\Rightarrow 6 p^{3} - 15 p^{2} + 12 p - 3 = 0$
$\Rightarrow (p - 1)^{2} (2 p - 1) = 0 \Rightarrow p = 0, 1 or \frac{1}{2}$

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The correct options are B 12 C 1 D 03C2 p2(1−p)+3C3p3= 5C3 p3(1−p)2+ 5C4 p4(1−p)+ 5C5 p53p2(1−p)+p3=10p3(1−p)2+5p4(1−p)+p5 ⇒p=0 and 10p(1+p2−2p)+5p2(1−p)+p3−3(1−p)−p=0 ⇒6p3−15p2+12p−3=0 ⇒(p−1)2(2p−1)=0⇒p=0, 1 or 12