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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+19 votes

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?

- Three green faces and four red faces.
- Four green faces and three red faces.
- Five green faces and two red faces.
- Six green faces and one red face

+34 votes

Best answer

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.

$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.

0

Correct me if I'm wrong, the ^{7}C_{x }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 ($\frac{4}{6}$)^{x}and the rest of the (7-x) have a Red face with a probability of ($\frac{2}{6}$ )^{(7-x)}.

Am I correct?

0

@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.

0

"with **four** green faces and **two** red faces is rolled"

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

0

→ 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

→ 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

0

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??

0

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??

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??

0

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

+16 votes

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.

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

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.

+6 votes

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.

+2

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.

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.

0 votes

0 votes

**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**.

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