GATE ECE 2018 | GA | Question: 9

Question

A cab was involved in a hit and run accident at night. You are given the following data about the cabs in the city and the accident.

1. 85% of cabs in the city are green and the remaining cabs are blue.
2. A witness identified the cab involved in the accident as blue.
3. It is known that a witness can conectly identify the cab colour only 80% of the tune.

Which of the following options is closest to the probability that the accident was caused by a blue cab?

• A. 12%
• B. 15%
• C. 41%
• D. 80%

Comments

What is the difference between the statements :

"It is known that a witness can correctly identify the cab colour only 80% of the time."

and

"It is known that a witness can correctly identify EACH cab colour only 80% of the time." ?

Answer 1

Probability that the cab is a green cab $=0.85$

Probability that the cab is a blue cab $=0.15$

The witness can correctly identify the cab colour only $80\%$ of the time

So, probability when the witness is correct means when the witness identifies blue cab $= 0.8$

& Probability when witness is wrong $= 0.2$

We know, $\bf{P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}}}$

The probability that the accident was caused by a blue cab $=\dfrac{0.15*0.8}{(0.15*0.8) +(0.85*0.2) }$

$\qquad= \dfrac{1.2}{2.9}$

$\qquad= 0.41$

$\qquad= 41\%$   

Option (C)

Comments

What is the difference between the statements :

"It is known that a witness can correctly identify the cab colour only 80% of the time."

and

"It is known that a witness can correctly identify EACH cab colour only 80% of the time." ?

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There is no difference between these two statements.

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@Sukanya Das

The two statements have different implications for the witness's ability to identify cab colors.

1. "It is known that a witness can correctly identify the cab colour only 80% of the time."

This statement suggests that across all cab color identifications, the witness is correct 80% of the time. It doesn't specify performance for individual colors. For example:

• The witness might be 90% accurate with blue cabs
• 70% accurate with yellow cabs
• 80% accurate with green cabs

The overall accuracy averages to 80%.

2. "It is known that a witness can correctly identify EACH cab colour only 80% of the time."

This statement implies that the witness has the same 80% accuracy rate for every individual cab color. So:

• 80% accurate with blue cabs
• 80% accurate with yellow cabs
• 80% accurate with green cabs
• And so on for any other colors

The key difference is that the second statement specifies consistent performance across all colors, while the first allows for variation in accuracy between colors as long as the overall average is 80%.

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Thank you

 

Answer 2

1. Let $B$ be the event that the cab was blue, and $W$ be the event that the witness identified the cab as blue.

2. We need to find $P(B|W)$, the probability that the cab was blue given that the witness identified it as blue.

3. Given data:

• $P(B) = 15\% = 0.15$ (probability of a cab being blue)
• $P(\text{not} \, B) = 85\% = 0.85$ (probability of a cab not being blue)
• $P(W|B) = 80\% = 0.8$ (probability of correct identification)
• $P(W|\text{not} \, B) = 20\% = 0.2$ (probability of incorrect identification)

4. Using Bayes' theorem:

     $P(B|W) = \dfrac{P(B \cap W)}{P(W)}$

    $P(B|W) = \dfrac{P(B \cap W)}{P(W \cap B) + P(W \cap \text{not} \, B)}$

    $P(B|W) = \dfrac{P(W|B) * P(B)}{P(W|B) * P(B) + P(W|\text{not} \, B) * P(\text{not} \, B)}$

5. Calculating:

    $P(B|W) = \dfrac{(0.8 * 0.15)}{(0.8 * 0.15) + (0.2 * 0.85)}$

    $P(B|W) = \dfrac{0.12}{(0.12 + 0.17)}$

    $P(B|W) = \dfrac{0.12}{0.29}$

    $P(B|W) \approx 0.4138 \approx 41.38\% \approx 41\%$

6. Therefore, the closest probability that the cab was blue given that the witness identified it as blue is $41\%$.

Answer 3

Probability that the cab is a green cab = 0.85

Probability that the cab is a blue cab = 0.15

The witness can correctly identify the cab colour only 80% of the time

So, probability when the witness is correct means when the witness identifies blue cab = 0.8

& Probability when witness is wrong = 0.2

We know, $P(E) = \dfrac{Numbe of favorable outcomes}{Number of all possible outcomes}$

The probability that the accident was caused by a blue cab = $\dfrac{(0.15)*(0.8)}{((0.15)*(0.8)) +((0.85)*(0.2)) }$

   = $\dfrac{1.2}{2.9}$

   = 0.41

   = 41%     option C)