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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 2 · Set 3 · Q37
Three persons viz. Amber, Bonzi and Comet are manufacturing cars which run on petrol and on battery as well. Their production share in the market is…
Three persons viz. Amber, Bonzi and Comet are manufacturing cars which run on petrol and on battery as well. Their production share in the market is $\displaystyle 60 \%, 30 \%$ and $\displaystyle 10 \%$ respectively. Of their respective production capacities, $\displaystyle 20 \%, 10 \%$ and $\displaystyle 5 \%$ cars respectively are electric (or battery operated) . Based on the above, answer the following:
(i)(a)
What is the probability that a randomly selected car is an electric car?
Marking-scheme solution
$\displaystyle 37$(i)(a)
Let A = Amber manufactures the car
B = Bonzi manufactures the car
C = Comet manufactures the car
E = The selected car is electric
$\displaystyle P(A) = \dfrac{60}{100},\; P(B) = \dfrac{30}{100},\; P(C) = \dfrac{10}{100}$
$\displaystyle P(E) = P(A) \times P\left(\dfrac{E}{A}\right) + P(B) \times P\left(\dfrac{E}{B}\right) + P(C) \times P\left(\dfrac{E}{C}\right)$
$\displaystyle = \dfrac{60}{100} \times \dfrac{20}{100} + \dfrac{30}{100} \times \dfrac{10}{100} + \dfrac{10}{100} \times \dfrac{5}{100}$
$\displaystyle = \dfrac{155}{1000}\ or\ \dfrac{31}{200}$
$\displaystyle 37$(i)(b)
Let A = Amber manufactures the car
B = Bonzi manufactures the car
C = Comet manufactures the car
E = The selected car is a petrol car
$\displaystyle P(A) = \dfrac{60}{100},\; P(B) = \dfrac{30}{100},\; P(C) = \dfrac{10}{100}$
$\displaystyle P(E) = P(A) \times P\left(\dfrac{E}{A}\right) + P(B) \times P\left(\dfrac{E}{B}\right) + P(C) \times P\left(\dfrac{E}{C}\right)$
$\displaystyle = \dfrac{60}{100} \times \dfrac{80}{100} + \dfrac{30}{100} \times \dfrac{90}{100} + \dfrac{10}{100} \times \dfrac{95}{100}$
$\displaystyle = \dfrac{845}{1000}\ or\ \dfrac{169}{200}$
$\displaystyle 37$(ii)
\[P\left(\frac{C}{E}\right) = \frac{P(C) \times P\left(\dfrac{E}{C}\right)}{P(E)} \]
\[= \frac{\dfrac{10}{100} \times \dfrac{5}{100}}{\dfrac{60}{100} \times \dfrac{20}{100} + \dfrac{30}{100} \times \dfrac{10}{100} + \dfrac{10}{100} \times \dfrac{5}{100}} \]
\[= \frac{\dfrac{50}{10000}}{\dfrac{1550}{10000}} = \frac{1}{31} \]
$\displaystyle 37$(iii)
\[P\left(\frac{A \text{ or } B}{E}\right) = 1 - P\left(\frac{C}{E}\right) = 1 - \frac{1}{31} = \frac{30}{31} \]
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