GATE CSE 2018 | Question: GA-10

Question

A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment?

• A. Three green faces and four red faces.
• B. Four green faces and three red faces.
• C. Five green faces and two red faces.
• D. Six green faces and one red face

Comments

The number of green faces is twice the number of red faces in the dice. So, green is obviously more likely to come up more. A eliminated.

B is too close. Eliminated.

D is too much out-of-proportion. Eliminated.

Only A seems reasonable.

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JashanArora

its C.

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 By Using Binomial Distribution 

In last you not need to do too much calculation that i did bcoz 

your can see that in every option answer is multiple of 2^10 if you eliminate this then 

A= 35 B = 3*35 =70  , C = 21*4 =84 , D = 7*8= 56

so answer is option C

Answer 1

Correct Option- (C)

To Find Expectation Of Green Face
Let, $G_i$ be the indicator random variable denoting that the green face shows up at the $i_th$ throw.
Therefore, $P(Gi) = 1.P(G_i=1) + 0.P(G_i=0) = P(G_i=1) = 4/6 = 2/3$

Let, $G$ be the random variable denoting the no. of green faces in 7 throws of the die.

$\therefore G = G_1 + G_2 + G_3 +\ ...\ + G_7$

$\therefore P(G) = P(G_1 + G_2 + \ ... \ + G_7)$
$\Rightarrow P(G) = P(G_1) + P(G_2) + \ ... \ + P(G_7)$ 
$\Rightarrow P(G) = (2/3) + (2/3) + ... 7 \ times$
$\Rightarrow P(G) = 14/3 = 4.667 \approx 5$

Therefore, expected no. of green faces in 7 throws of the die is 5.

Similarly, proceed to find the expected no. of red faces, which comes out to be 2.
 

Answer 2

Answer is option C.