$\binom{3}{2}*(\frac{1}{3})^{2}* \frac{2}{3} + \binom{3}{3}*(\frac{1}{3})^{3} = \frac{7}{27} = 0.259$

4 votes

Consider an unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice. If the dice is thrown thrice, the probability of obtaining red colour on top face of the dice at least twice is----

I am getting 0.75 can anyone confirm this ?

I am getting 0.75 can anyone confirm this ?

3 votes

Best answer

0.2592

Unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice means $\frac{2}{6} = \frac{1}{3}$ is probability of each color comes on top.

Now ask Atleast two

For 2= $\binom{3}{2}(\frac{1}{3})^{2} \frac{2}{3}$ = $\frac{4}{9}$

For 3= $\binom{3}{3}(\frac{1}{3})^{3}$ = $\frac{1}{27}$

Total = $\frac{7}{27}= 0.252$