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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?

1. Three green faces and four red faces.
2. Four green faces and three red faces.
3. Five green faces and two red faces.
4. Six green faces and one red face

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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.

We can calculate the probability of each of the options but by logic we can easily eliminate the ones with more number of ${\color{Red} {red}}$ faces - option $A$ can be avoided without any calculation.

$A = P(3 {\color{Green} G},4{\color{Red} R}) = {}^7C_3. {\color{Green}{\left(\frac{4}{6}\right)^3}}{\color{Red}{\left(\frac{2}{6}\right)^4}}$

$B = P(4 {\color{Green} G},3{\color{Red} R}) = {}^7C_4. {\color{Green}{\left(\frac{4}{6}\right)^4}}{\color{Red}{\left(\frac{2}{6}\right)^3}}$

$C = P(5 {\color{Green} G},2{\color{Red} R}) = {}^7C_5. {\color{Green} {\left(\frac{4}{6}\right)^5}}{\color{Red} {\left(\frac{2}{6}\right)^2}}$

$D = P(6 {\color{Green} G},1{\color{Red} R}) = {}^7C_6.{\color{Green} { \left(\frac{4}{6}\right)^6}}{\color{Red}{\left(\frac{2}{6}\right)^1}}$

Option $A$ is clearly smaller and hence eliminated.

$\frac{C}{B} = \frac{{}^7C_5}{{}^7C_4}.\frac{4}{6}. \frac{6}{2} = \frac{3}{5} . 2 > 1 .$

So, $C > B.$

$\frac{C}{D} = \frac{{}^7C_5}{{}^7C_6}.\frac{6}{4}. \frac{2}{6} = \frac{21}{7}.0.5 > 1 .$

So, $C > D.$

Hence, $C$ is the most favourable outcome.
by Veteran (431k points)
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C/B=(7C5/* 7C3) *4/6*2/6

**2/6
+3
divided by 2/6 = multiply by 6/2
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Correct me if I'm wrong, the 7Cx part for every option corresponds to choosing x of the 7 throws, such that those x throws have a Green face with a probability of ($\frac{4}{6}$)xand the rest of the (7-x) have a Red face with a probability of ($\frac{2}{6}$ )(7-x).

Am I correct?

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@Arjun Please share links to some easy (but equally good) lecture notes on probability to suffice for the GATE exams as of now. I am very weak in probability. I am looking for something that could be intuitively understood, instead of going through complicated formulas.

I had a hard time solving the given problem, and I failed. When I looked at the answer, I didn't understood from where the $\binom{n}{x}$ came from.

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why option A) will be avoided?
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"with four green faces and two red faces is rolled"

So, more red faces than green faces is clearly not a favorable case.

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→ Each side of a unbiased die can have equal probability i.e., = 1/6
→ If we roll a die for six time then we get 4 green faces and 2 red faces.
→ And if we roll for seventh time green face can have more probability to become a outcome.
→ Then most likely outcome is five green faces and two red faces
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If we roll a die for six time then we get 4 green faces and 2 red faces.

When die is rolled, each face can come in upper face of die with probability $\frac{1}{6}.$ Practically, when dice rolled $7$ times, any faces can be a top face.

Isn't it??

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For elimination just arrange the values of all cases in descending order and pick the first one.
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why arrange in descending order??

See in exam hall, many probability comes in mind.

So, u have to pick best.

If u go with above logic of @Arjun Sir

Then A) and B) both have equal probability to be ans.

Is it not??
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It is not necessary to arrange in desc order. I told so that we can get the highest value. please see my answer @srestha

Expectation of green face in a single throw

$E[Green] = 1*(\frac{4}{6}) + 0*(1-\frac{4}{6}) = \frac{4}{6}$

Expectation of red face in a single throw

$E[Red] = 1*(\frac{2}{6}) + 0*(1-\frac{2}{6}) = \frac{2}{6}$

now in $7$ throws $\rightarrow \ E[Green] = 7*(\frac{4}{6}) = \frac{4}{6} = 4.66 \approx 5$

now in $7$ throws $\rightarrow \ E[Red] = 7*(\frac{2}{6}) = \frac{14}{6} = 2.33 \approx 2$

$\therefore$ Most likely we get $5$ green and $2$ red faces.

So, option $C.$ is the corrrect answer.
by Active (1.6k points)
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is this approach  work for every problem?
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It can work where we have to find mean value

but sometimes problem are not direct like this one where this approach work , so we have to comprehend it using keywords here is 'likely'
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ok....but

what is mean by exactly "mean value" .

if you think average then here you direct calculated seven thrown.

Which of the following combinations is the most likely outcome of the experiment ?

here most likely means that we have to pick the options which has the highest value so we can compare the options and select the appropriate one.

$A = P(3 {\color{Green} G},4{\color{Red} R}) = {}^7C_3. {\color{Green}{\left(\frac{4}{6}\right)^3}}{\color{Red}{\left(\frac{2}{6}\right)^4}}$

$B = P(4 {\color{Green} G},3{\color{Red} R}) = {}^7C_4. {\color{Green}{\left(\frac{4}{6}\right)^4}}{\color{Red}{\left(\frac{2}{6}\right)^3}}$

$C = P(5 {\color{Green} G},2{\color{Red} R}) = {}^7C_5. {\color{Green} {\left(\frac{4}{6}\right)^5}}{\color{Red} {\left(\frac{2}{6}\right)^2}}$

$D = P(6 {\color{Green} G},1{\color{Red} R}) = {}^7C_6.{\color{Green} { \left(\frac{4}{6}\right)^6}}{\color{Red}{\left(\frac{2}{6}\right)^1}}$

since we are comparing so multiplying each option with $6^{7}$ will not change the result.

$A = {}^7C_3. {\color{Green}{\left(\frac{4}{1}\right)^3}}{\color{Red}{\left(\frac{2}{1}\right)^4}}$

$B = {}^7C_4. {\color{Green}{\left(\frac{4}{1}\right)^4}}{\color{Red}{\left(\frac{2}{1}\right)^3}}$

$C = {}^7C_5. {\color{Green} {\left(\frac{4}{1}\right)^5}}{\color{Red} {\left(\frac{2}{1}\right)^2}}$

$D = {}^7C_6.{\color{Green} { \left(\frac{4}{1}\right)^6}}{\color{Red}{\left(\frac{2}{1}\right)^1}}$

Now convert each of the ${\color{Green} {4^{x}}}$ term into ${\color{Green} {2^{2x}}}$ term and multiply them with ${\color{Red} {2^{y}}}$

$A = {}^7C_3. 2^{10}$

$B = {}^7C_4. 2^{11}$

$C = {}^7C_5. 2^{12}$

$D = {}^7C_6. 2^{13}$

since we are comparing so dividing each option with $2^{10}$ will not change the result.

$A = {}^7C_3.=35$

$B = {}^7C_4. 2=70$

$C = {}^7C_5. 2^{2}=84$

$D = {}^7C_6. 2^{3}=56$

$\because$ Option $C$ has the highest value so it is the correct answer.

$\therefore$ option $C$ is the correct answer.

by Boss (24k points)
edited by
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No, answer is C according to official key.
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Typing mistake
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@Satbir Why binomial distribution is applied here?

+1
It is not actually binomial , we are just selecting the balls and multiplying them with their respective probability.

you can think it as binomial since there are only 2 cases and whichever ball we select will fall in these 2 cases :-

p = prob of success = getting green ball

q = prob of failure = not getting green ball = getting red ball.

if we would have given more than 2 like 3 colours then this binomial trick cant be applied bcoz there will be 3 cases.
Apply Binomial theorem on all 4. C will be having the highest probability among all.

Ans: C
by Active (1.3k points)
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plz explain
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By applying binomial theorem probability of c is greater than d. C probability coming out to be .24 and D probability coming out to be .20

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.

by Junior (831 points)