retagged by
14,654 views
30 votes
30 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?

  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
retagged by

6 Answers

Best answer
54 votes
54 votes
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.
edited by
23 votes
23 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.
edited by
10 votes
10 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.

edited by
2 votes
2 votes
Apply Binomial theorem on all 4. C will be having the highest probability among all.

Ans: C
Answer:

Related questions

29 votes
29 votes
4 answers
1
gatecse asked Feb 14, 2018
10,591 views
In the figure below, $\angle DEC + \angle BFC$ is equal to _____$\angle BCD - \angle BAD$$\angle BAD + \angle BCF$$\angle BAD + \angle BCD$$\angle CBA + \angle ADC$
15 votes
15 votes
10 answers
4
gatecse asked Feb 14, 2018
7,305 views
If $pqr \ne 0$ and $p^{-x}=\dfrac{1}{q},q^{-y}=\dfrac{1}{r},r^{-z}=\dfrac{1}{p},$ what is the value of the product $xyz$ ?$-1$$\dfrac{1}{pqr}$$1$$pqr$