GATE 2020 EC Q.54: solution and answer

GATE 2020 · ECQ.54 · 2 marks · NAT

$X$ is a random variable with uniform probability density function in the interval $[- 2, 10]$ . For $Y = 2 X - 6$ , the conditional probability $P (Y \leq 7 ∣ X \geq 5)$ (rounded off to three decimal places) is ________.

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Answer and solution

ANSWER

$P (Y \leq 7 ∣ X \geq 5) = 0.300$

Official answer: 0.299 to 0.301Source: Official answer key, GATE 2020 EC (IIT Delhi)

Write the PDF of $X$ . For a uniform distribution on $[- 2, 10]$ , $f_{X} (x) = \frac{1}{10 - ( - 2 )} = \frac{1}{12}$ for $- 2 \leq x \leq 10$ , and $0$ everywhere else.
Rewrite the event on $Y$ as an event on $X$ . Since $Y = 2 X - 6$ is strictly increasing, $Y \leq 7 ⟺ 2 X - 6 \leq 7 ⟺ 2 X \leq 13 ⟺ X \leq 6.5$ .
Use the definition of conditional probability: $P (Y \leq 7 ∣ X \geq 5) = \frac{P ( X \leq 6.5 \cap X \geq 5 )}{P ( X \geq 5 )} = \frac{P ( 5 \leq X \leq 6.5 )}{P ( 5 \leq X \leq 10 )}$ (the upper limit is $10$ because $X$ never exceeds $10$ ).
For a uniform variable, the probability of a sub-interval is its length times the density $\frac{1}{12}$ : $P (5 \leq X \leq 6.5) = \frac{1.5}{12}$ and $P (5 \leq X \leq 10) = \frac{5}{12}$ .
Divide: $P (Y \leq 7 ∣ X \geq 5) = \frac{1.5/12}{5/12} = \frac{1.5}{5} = 0.300$
Check: once $X \geq 5$ is known, $X$ is uniform on $[5, 10]$ (length $5$ ). The part of that interval with $X \leq 6.5$ has length $1.5$ , so the probability is $1.5/5 = 0.3$ .

KEY IDEA

Turn the condition on $Y$ into a condition on $X$ using the increasing map $Y = 2 X - 6$ , which gives $X \leq 6.5$ . Conditioning a uniform variable on a sub-interval leaves it uniform on that sub-interval, so the conditional probability is just a ratio of lengths: $1.5/5$ .

WATCH OUT FOR

Dividing by the full length $12$ and not by the length of the conditioning interval $[5, 10]$ , which gives $1.5/12 = 0.125$ . Other slips: solving $2 X - 6 \leq 7$ as $X \leq 0.5$ or $X \leq 6$ ; forgetting to cap $X \geq 5$ at the support limit $10$ ; or computing $P (Y \leq 7)$ over the whole range, $8.5/12 \approx 0.708$ , and never applying the condition.

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GATE 2020 EC Q.54: solution and answer

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