GATE 2024 DA Q.65: solution and answer

GATE 2024 · DAQ.65 · 2 marks · NAT

Two fair coins are tossed independently. $X$ is a random variable that takes a value of 1 if both tosses are heads and 0 otherwise. $Y$ is a random variable that takes a value of 1 if at least one of the tosses is heads and 0 otherwise.

The value of the covariance of $X$ and $Y$ is ______ (rounded off to three decimal places).

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Answer and solution

ANSWER

The covariance of $X$ and $Y$ is $\frac{1}{16} = 0.0625$ , which is about $0.063$ to three decimal places.

Official answer: 0.062 to 0.063Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

Two fair coins give four equally likely outcomes: HH, HT, TH, TT, each with probability $\frac{1}{4}$ .
$X = 1$ only for HH, so $E [X] = P (X = 1) = \frac{1}{4}$ .
$Y = 0$ only for TT, so $E [Y] = P (Y = 1) = 1 - \frac{1}{4} = \frac{3}{4}$ .
$X Y = 1$ only when both $X = 1$ and $Y = 1$ . Outcome HH has $X = 1$ and also $Y = 1$ (it has at least one head), so $X Y = 1$ exactly when the outcome is HH. Hence $E [X Y] = P (H H) = \frac{1}{4}$ .
Use $Cov (X, Y) = E [X Y] - E [X] E [Y]$ .
$Cov (X, Y) = \frac{1}{4} - \frac{1}{4} \cdot \frac{3}{4} = \frac{1}{4} - \frac{3}{16} = \frac{1}{16} = 0.0625.$
The exact value is $0.0625$ (rounded to three decimals it is about $0.063$ ).

KEY IDEA

Write each indicator's expectation as a probability, then use $Cov (X, Y) = E [X Y] - E [X] E [Y]$ . Because $X = 1$ forces $Y = 1$ , the product $X Y$ equals $X$ .

WATCH OUT FOR

Treating $X$ and $Y$ as independent and answering $0$ . They are dependent, since "both heads" is a special case of "at least one head". Another slip is computing $E [X Y]$ as $E [X] E [Y]$ .

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 4 Descriptive statistics questions, with notes

GATE 2024 DA Q.65: solution and answer

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