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

Answer 1

The correct option is C

Five green faces and two red faces
Four green, two red face

$P\left(G\right)=\frac{4}{6}=\frac{2}{3}q\left(R\right)=\frac{1}{3}n=7\text{Option (1), P(G = 3)}{=}^7{C}_3{\left(\frac{2}{3}\right)}^3{\left(\frac{1}{3}\right)}^4=\frac{35\times 2^3}{(3{)}^7}=\frac{35\times 2^3}{(3{)}^7}\left(2\right),P(G=4){=}^7{C}_4\times {\left(\frac{2}{3}\right)}^4\times {\left(\frac{1}{3}\right)}^3=\frac{35\times 2^4}{(3{)}^7}=\frac{35\times 2^4}{(3{)}^7}\left(3\right),P(G=5){=}^7{C}_5\times {\left(\frac{2}{3}\right)}^5\times {\left(\frac{1}{3}\right)}^2=\frac{21\times 2^5}{(3{)}^7}=\frac{42\times 2^4}{(3{)}^7}\left(4\right),P(G=6){=}^7{C}_6\times {\left(\frac{2}{3}\right)}^6\times \left(\frac{1}{3}\right)=\frac{7\times 2^6}{(3{)}^7}=\frac{28\times 2^4}{(3{)}^7}$

Option 3 is maximum value.

So, five green faces and two red faces.